Mathematics Tricks & Tips of Permutation and Combination for NDA
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Tricks & Tips of Permutation and Combination for NDA

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Fundamental Principle of Counting and factorial

Q 2117601589

What is the number of ways in which `3` holiday travel
tickets are to be given to `10` employees of an
organisation, if each employee is eligible for any one or
more of the tickets?
NDA Paper 1 2016
(A)

`60`

(B)

`120`

(C)

`500`

(D)

`1000`

Solution:

Since, each ticket can be given to any one of `10`

employees of an organisation.

`:.` Required number of ways

`= 10 xx 10 xx 10 = 10^(3) = 1000`
Correct Answer is `=>` (D) `1000`
Q 1669591415

What is the number of ways in which one can post
`5` letters in `7` letter boxes?
NDA Paper 1 2014
(A)

`7^5`

(B)

`3^5`

(C)

`5^7`

(D)

`2520`

Solution:

There are `5` letters and `7` letter boxes.

First letter can be put any `7` letters boxes = `7` ways

Similarly, `2nd, 3rd, 4th` and `5th` letters be put in `7` ways

each, respectively.

`:. 7 xx 7 xx 7 xx 7 xx 7 = 7^5`
Correct Answer is `=>` (A) `7^5`
Q 2339512412

If there are `4` candidates for the post of a lecturer
in Mathematics and one is to be selected by votes
of `5` men, then what is the number of ways in
which the votes can be given?
NDA Paper 1 2011
(A)

`1048`

(B)

`1072`

(C)

`1024`

(D)

`625`

Solution:

A man can give votes for `4` candidates `= 4` ways

`=> 5` men can give votes for `4` candidates

`= 4 xx 4 xx 4 xx 4 xx 4` ways

`:.` Required number of ways `= (4)^5 = 1024` ways
Correct Answer is `=>` (C) `1024`
Q 2339145012

On a railway route there are `20` stations. What is
the number of different tickets required in order
that it may be possible to travel from every station
to every other station?
NDA Paper 1 2007
(A)

`40`

(B)

`380`

(C)

`400`

(D)

`420`

Solution:

Required number of tickets `= 20 xx 19 = 380`
Correct Answer is `=>` (B) `380`
Q 2479645516

There are `n` different books and `p` copies of each. The number of ways in which a selection can be made from them, is
BCECE Stage 1 2011
(A)

`n^p`

(B)

`p^n`

(C)

`(p+1)^n -1`

(D)

`(n+1)^p-1`

Solution:

In case of each book we may take `0, 1, 2, 3, ... , p` copies, i.e., we may deal with each book in `(p + 1)` ways and therefore with all the books in `(p + 1)^n` ways. But this includes the case where all the books rejected and no selection is made.

So, the number of ways in which selection can be made `= (p + 1)^n-1`
Correct Answer is `=>` (C) `(p+1)^n -1`
Q 2349123013

How many diagonals will be there in an n-sided
regular polygon?
NDA Paper 1 2011
(A)

`(n(n - 1))/2`

(B)

`(n(n - 3))/2`

(C)

`n^2 - n`

(D)

`(n(n + 1))/2`

Solution:

Total number of lines in an n-sided regular polygon

`= text()^nC_2`

and total number of sides in n-sided regular polygon `= n`

`:.` Number of diagonals in n-sided regular polyuon `= text()^nC_2 - n`

`= (n(n- 1))/n - n = n ((n- 1)/2 - 1) = (n(n- 3))/2`
Correct Answer is `=>` (B) `(n(n - 3))/2`
Q 2561191925

Four speakers will address a meeting where speaker Q will always speak P. Then, the number of ways in which the order of speakers can be prepared is
WBJEE 2012
(A)

`256`

(B)

`128`

(C)

`24`

(D)

`12`

Solution:

Four speakers will address the meeting in

`4!` ways =` 24` different ways in which half

number of cases will be such that P speaks

before `Q` and half number of case will be such

that `P` speaks after `Q`

`:.` Required number of ways ` = (24)/2 = 12`
Correct Answer is `=>` (D) `12`
Q 2086378277

Three letters can be posted in five letter boxes in ......... ways.
NCERT Exemplar
(A) True
(B) False
Solution:

Required number of ways `= 5^3 = 125`
Correct Answer is `=>` (B)
Q 2046478373

In a steamer there are stalls for `12` animals and there are horses, cows
and calves (not less than `12` each) ready to be shipped. They can be
loaded in `3^(12)` ways.
NCERT Exemplar
(A) True
(B) False
Solution:

There are three types of animals and stalls available for `12` animals.

Number of ways of loading `= 3^(12)`
Correct Answer is `=>` (A)

Combination : As a selection with repetition and without repetition

`text()^(n)C_r = (n!)/(r! (n-r)! )`
Q 2280391217

The number of ways in which a cricket team of `11` players
be chosen out of a batch of `15` players, so that the captain of
the team is always included, is
NDA Paper 1 2015
(A)

`165`

(B)

`364`

(C)

`1001`

(D)

`1365`

Solution:

Total number of selection of `11` players out of

`15` players in which captain is included `= text()^(14)C_(10)`

` = (14!)/(10! 4!) = (14 xx 13 xx 12 xx 11)/(1xx2 xx3 xx4) = 1001`
Correct Answer is `=>` (C) `1001`
Q 1668623505

What is ` sum_(r = 0)^1 text()^(n+r)C_n`, equal to?
NDA Paper 1 2015
(A)

`text()^(n+2)C_1`

(B)

`text()^(n+2)C_n`

(C)

`text()^(n+3)C_n`

(D)

`text()^(n+2)C_(n+1)`

Solution:

` sum_(r = 0)^1 text()^(n+r)C_n = sum_(r = 0)^1 text()^(n+r)C_(n+r-n) = sum_(r = 0)^1 text()^(n+r)C_r`

` = text()^(n)C_0 + text()^(n+1)C_1 = 1 + (n+1!)/(n!1!)`

` = 1 + n + 1 = n + 2 = text()^(n+2)C_1`
Correct Answer is `=>` (A) `text()^(n+2)C_1`
Q 1679591416

What is the number of ways that a cricket team of
`11` players can be made out of `15` players?
NDA Paper 1 2014
(A)

`364`

(B)

`1001`

(C)

`1365`

(D)

`32760`

Solution:

`11` players can be made out of `15` players is

` text()^(15)C_(11) = (15!)/(11 ! 4!)`

` = (15 xx 14 xx 13 xx 12 xx 11!)/ (11! xx 1 xx 2 xx 3 xx 4)`

`= 1365`
Correct Answer is `=>` (C) `1365`
Q 1713167949

Out of `7` consonants and `4` vowels, words are to be
formed by involving `3` consonants and `2` vowels. The
number of such words formed is
NDA Paper 1 2014
(A)

`25200`

(B)

`22500`

(C)

`10080`

(D)

`5040`

Solution:

In out of `7` consonants, `3` consonants can be

selected in `text()^(7)C_3` ways.

In out of `4` vowels, `2` vowels can be selected in `text()^(4)C_2` ways.

:. Number of such words `= text()^(7)C_3 xx text()^(7)C^2 xx 5!`

(since, `5` letters can be selected in `5` ways)

` = 35 xx 6 xx 120`

` = 25200`
Correct Answer is `=>` (A) `25200`
Q 2329323211

`5` books are to be chosen from a lot of `10` books. If
m is the number of ways of choice when one
specified book is always included and n is the
number of ways of choice when a specified book
is always excluded, then which one of the
following is correct?
NDA Paper 1 2011
(A)

m > n

(B)

m = n

(C)

m = n -1

(D)

m = n - 2

Solution:

Number of ways when one specified book is included

`= text()^9C_4 = m => m = 126`

and number of ways when one specified book is excluded

`= text()^9C_5 = n`

`=> n = 126 => m = n`
Correct Answer is `=>` (B) m = n
Q 2339123912

How many times does the digit `3` appear while
writing the integers from `1` to `1000`?
NDA Paper 1 2009
(A)

`269`

(B)

`271`

(C)

`300`

(D)

None of these

Solution:

Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

Case 1. The numbers in which 3 occurs only once. This means that 3 is one of the digits and the remaining two digits will be any of the other 9 digits

You have 1*9*9 = 81 such numbers. However, 3 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 3 will appear only once.

Case 2. The numbers in which 3 will appear twice. In these numbers, one of the digits is not 3 and it can be any of the 9 digits.
There will be 9 such numbers. However, this digit which is not 3 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 3 is written twice. Therefore, 3 is written 54 times.

Case 3. The number in which 3 appears thrice - 333 - 1 number. 3 is written thrice in it.

Therefore, the total number of times the digit 3 is written between 1 and 999 is 243 + 54 + 3 = 300
Correct Answer is `=>` (C) `300`
Q 2865401365

We have to choose Eleven players for cricket team from eight batsmen, six bowlers, four all rounder and two wicket keepers in the following conditions.
The number of selections, when atmost one all rounder and one wicket keeper will play

(A)

`text()^4C_1 xx ^14C_10 + ^2 C_1 xx ^14C_10 + ^4C_1xx^2C_1 xx ^14C_9 + ^14C_11`

(B)

`text()^4C_1 xx^15C_11 + ^15C_11`

(C)

`text()^4C_1 xx ^15C_10 + ^15C_11`

(D)

None of the above

Solution:

When one all rounder and ten players from bowlers and batsmen play, number of ways is `text()^4C_1 xx ^14C_10`

When, one wicketkeeper and 10 players from bowlers and batsmen play, number of ways is `text()^2C_1 xx ^14C_10`

When one all rounder, one wickctkeeper and nine players from batsmen and bowlers play, number of ways is `text()^4C_1 xx ^2C_! xx ^4 C_9`

When all eleven players play from bowlers and batsman, then number of ways is `text()^14C_11`

Total number of selections is

`text()^4C_1 xx text()^14C_10 + ^2 C_1 xx ^14C_10 + ^4C_1 xx ^2C_1 xx ^14C_9 + ^14C_11`
Correct Answer is `=>` (A) `text()^4C_1 xx ^14C_10 + ^2 C_1 xx ^14C_10 + ^4C_1xx^2C_1 xx ^14C_9 + ^14C_11`
Q 2815201169

There are ten points in a plane, of these ten points four points are in a straight line and with the exception of these four points, no three points are in the same straight line. On the basis of this information answer the following questions.
The number of triangles formed by joining these ten points is

(A)

116

(B)

80

(C)

96

(D)

56

Solution:

Total number of selections of 3 points out of 10 points

`= text()^10C_3 = (10 xx 9 xx 8 ) /( 3!) =120`

When, 3 collinear points are selected no triangle is formed.

Number of selections of 3 points out of 4 collinear points

`= text()^4C_3 = 4`

`:. ` Required number `= 120- 4 = 116`
Correct Answer is `=>` (A) 116
Q 2845101063

A committee of five members is to be made from 4 gentlemen and 6 ladies. Then, find the number of ways in which committee can be formed.
Gentlemen are in majority

(A)

60

(B)

66

(C)

120

(D)

126

Solution:

Number of ways to make the committee, when gentlemen arc in majority

`= text()^4C_3 text()^6C_2 + text()^4C_4 text()^6 C_1`

`= 4 xx 15 + 1 xx 6 = 66`
Correct Answer is `=>` (B) 66
Q 2319723610

In how many ways can a committee consisting of
`3` men and `2` women be formed from `7` men and
`5` women?
NDA Paper 1 2010
(A)

45

(B)

350

(C)

700

(D)

4200

Solution:

Required number of ways

`= text()^7C_3 xx text()^5C_2 = (7 xx 6 xx 5)/(3 xx 2) xx (5 xx 4)/(2 xx 1) = 35 xx 10 = 350`
Correct Answer is `=>` (B) 350
Q 2389523417

`A, B, C, D` and `E` are coplanar points and three of
them lie in a straight line. What is the maximum
number of triangles that can be drawn with these
points as their vertices?
NDA Paper 1 2011
(A)

5

(B)

9

(C)

10

(D)

12

Solution:

Number of triangles using `5` points out of three are on a

straight line

`= text()^5C_3 - text()^3C_3 = (5 !)/( 3! 2!) - 1`

` = (5 xx 4)/2 - 1 = 10 - 1 = 9`
Correct Answer is `=>` (B) 9
Q 2319823710

A team of `8` players is to be chosen' from a group of
`12` players. Out of the eight players one is to be
elected as captain and another an vice-captain. In
how many ways can this is done?
NDA Paper 1 2010
(A)

27720

(B)

13860

(C)

6930

(D)

495

Solution:

Number of ways to choose `8` players from `12` players

`= text()^(12)C_8 = (12!)/(8! 4!) = 495`

and number of ways to choose a captain and a vice-captain

`= text()^8C_1 xx text()^7C_1 = 8 xx 7 = 56`

`:.` Required number of ways `= 495 xx 56 = 27720`
Correct Answer is `=>` (A) 27720
Q 2379334216

From `7` men and `4` women a committee of `6` is to
be formed such that the committee contains at
least two women. What is the number of ways to
do this?
NDA Paper 1 2008
(A)

`210`

(B)

`371`

(C)

`462`

(D)

`5544`

Solution:

The number of ways

`= text()^(11)C_6 - ( text()^7C_6 xx text()^4C_0 + text()^7C_5 xx text()^4C_1)`

` = (11 xx 10 xx 9 xx 8 xx 7) /(5 xx 4 xx 3 xx 2) - ( 7 + ( 7 xx 6)/2 xx 4)`

`= 462 - (7 + 84) = 371`
Correct Answer is `=>` (B) `371`
Q 2309634518

A group consists of `5` men and `5` women. If the
number of different five person committees
containing `k` men and `(5 - k)` women is `100`, what
is the value of `k`?
NDA Paper 1 2008
(A)

Only 2

(B)

Only 3

(C)

2 or 3

(D)

Only 4

Solution:

` text()^5C_k xx text()^5C_(5 - k) = 100`

` => (5!)/ (k ! (5 - k)!) xx (5!)/( (5- k)! 5 !) = 100`

` => ( (5 !)/( k ! (5 - k)!))^2 = 100`

`:. (5 !)/( k ! (5 - k)!) = text()^5C_k = 100`

which is true for ` k = 2` or `3`.
Correct Answer is `=>` (C) 2 or 3

Permutations : As a arrangement with repetition and without Repetition

`text()^(n)P_r = (n!)/( (n-r)! ) = text()^(n)C_r .r!`
Q 2733780642

Three-digit numbers are formed from the digits `1, 2` and `3` in such a way that the digits are not repeated. What is the sum of such three-digit numbers?
NDA Paper 1 2017
(A)

`1233`

(B)

`1322`

(C)

`1323`

(D)

`1332`

Solution:

Three digits Number `tt(( 1, 2, 3),(1, 3, 2),(2, 1, 3), (2, 3, 1), (3, 1, 2), ( 3, 2, 1))`

Sum of all digits `= 1332`
Correct Answer is `=>` (D) `1332`
Q 2733180942

The number of different words (eight letter words) ending and beginning with a consonant which can be made out of the letters of the word 'EQUATION' is
NDA Paper 1 2017
(A)

`5200`

(B)

`4320`

(C)

`3000`

(D)

`2160`

Solution:

First and last letters can be selected from `4` consonants in `text()^4 C_2` ways `= 6` ways

Remaining `6` letters cab be arranged in `6` places in `6!` ways.

Total different words `= 720 xx 6 = 4320`
Correct Answer is `=>` (B) `4320`
Q 2701256128

A five -digit number divisible by 3 is to be formed using the digits `0,1,2,3` and `4` without repetition of digits. what is the number of ways this can be done ?
NDA Paper 1 2016
(A)

96

(B)

48

(C)

32

(D)

No Number can be formed

Solution:

For a Number divisible by `3` the sum of the digits must be divisible by `3`

As the sum of the given number `(0+1+2+3+4=10)` is not divisible by `3`

`:.` No number can be performed .
Correct Answer is `=>` (D) No Number can be formed
Q 2137101982

What is the number of four-digit decimal number `(< 1 )`
in which no digit is repeated ?
NDA Paper 1 2016
(A)

`3024`

(B)

`4536`

(C)

`5040`

(D)

None of these

Solution:

Clearly, number of four-digit decimal numbers that

can be formed using the digits `0, 1, 2, 3, 4, 5, 6, 7, 8, 9`,

when no digit is repeated, are given by

` text ()^(10)P_(4) - text ()^(9)P_(3) = (10!)/(6!) - (9!)/(6!) = (9!)/(6!) (10 -1)`

` = (9 xx 8 xx7 xx6! xx 9) /(6!) = 72 xx 7 xx 9 = 4536`
Correct Answer is `=>` (B) `4536`
Q 2210291119

If different words are formed with all the letters of the
word `'AGAIN'` and are arranged alphabetically among
themselves as in a dictionary, the word at the `50`th place
will be
NDA Paper 1 2015
(A)

NAAGI

(B)

NAAIG

(C)

IAAGN

(D)

IAANG

Solution:

We have, word 'AGAIN'

The letter starts from `A AGIN = 4! = 24`

The letter starts from `GA AIN = (4!)/(2!) = 12`

The letters starts from `IA AGN = (4!)/(2!) = 12`

Number of total letters `= 48`

`:. 49`th word is `NA AGI` and 50th word is `NA AIG`.
Correct Answer is `=>` (B) NAAIG
Q 2281112927

The number of `3`-digit even numbers that can be formed
from the digits `0, 1, 2, 3, 4` and `5`, repetition of digits being
not allowed, is
NDA Paper 1 2015
(A)

`60`

(B)

`56`

(C)

`52`

(D)

`48`

Solution:

We have, digits `0, 1, 2, 3, 4` and `5`

[even number]

(i) When `0` is at unit place, then

` tt ((xx,xx,xx),(4,5,0)) = 20` numbers

(ii) When 0 is not at unit place, then

` tt ((4,4,2),(xx,xx,xx)) = 32` numbers

`2` or `4`

`:.` Total even numbers `= 20 + 32 = 52`
Correct Answer is `=>` (C) `52`
Q 2339312212

The number of permutations that can be formed
from all the letters of the word 'BASEBALL: is
NDA Paper 1 2012
(A)

`540`

(B)

`1260`

(C)

`3780`

(D)

`5040`

Solution:

These are `8` letters in word 'BASEBALL' in which `2B, 2A,`

`2L, 1S` and `1E`.

So, the number of permutations that can be formed from all the

letters of the word 'BASEBALL'.

` = (8!)/(2!2!2!) = ( 8 xx 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1)/( 2 xx 1 xx 2 xx 1 xx 2 xx 1)`

`= 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1`

`= 42 xx 120 = 5040`
Correct Answer is `=>` (D) `5040`
Q 2781767627

What is the number of odd integers between 100 and 999 with no digits repeated ?
NDA Paper 1 2016
(A)

2100

(B)

2120

(C)

2240

(D)

3331

Solution:

you'll have to place 1,3,5,7,9 in unit place to make number odd.

So in unit place No. of choice `=5 { 1,3,5,7,9}`

In thousand place no. of choice ` =8 ` (0 left out & No reception)

Similarly `8 & 7` choices in hundreds & less place

total No, of choices `= 8 xx 8 xx 7 xx 5 = 2240`
Correct Answer is `=>` (C) 2240
Q 1688623507

How many words can be formed using all the
letters of the word 'NATION, so that all the three
vowels should never come together?
NDA Paper 1 2015
(A)

`354`

(B)

`348`

(C)

`288`

(D)

None of these

Solution:

Required number of words

`= (6 !)/(2!) - ( 4! xx 3!)/(2!) = 360 - 72 = 288`
Correct Answer is `=>` (C) `288`
Q 2512180939

The number of permutations by taking all letters and keeping the vowels of the word COMBINE in the odd places is
WBJEE 2010
(A)

`96`

(B)

`144`

(C)

`512`

(D)

`576`

Solution:

Vowels are O.I.E. {In the required word}
No. of odd place = 4

`therefore` No. of ways = `text()^(4)P_3xx4! = 576`
Correct Answer is `=>` (D) `576`
Q 2551891724

A vehicle registration number consists of 2 letters of English alphabet followed by 4 digits, where the first digit is not zero. Then, the total number of vehicles with distinct registration numbers is
WBJEE 2012
(A)

`26^2 xx 10^4`

(B)

`text()^(26)P_2 xx text()^(10)P_4`

(C)

`text()^(26)P_2 xx 9 xx text()^(10)P_3`

(D)

`26^2 xx 9 xx 10^3`

Solution:

The total number of arrangements of 2 letters of

English alphabet

`= 26 xx 26`

The total number of arrangements of `4` digits

number in which first digit is not zero

`= 9 xx 10 xx 10 xx 10`

`:.` The total number of vehicles with distinct

registration number

`= 26 xx 26 xx 9 xx 10 xx 10 xx 10`

`= 26^2 xx 9 xx 10^3`
Correct Answer is `=>` (D) `26^2 xx 9 xx 10^3`
Q 2814691559

Find the number of numbers between 400 and 4000 that can be formed with the digits 2, 3, 4, 5, 6 and 0.
When number is of 3 digits

(A)

75

(B)

90

(C)

60

(D)

80

Solution:

Since, the number should be greater than 400 therefore, hundreds place can be filled up by any one of the three digits 4, 5, 6 in 3 different ways.
Remaining two places can be filled up by remaining five digits in `text()^5P_2` ways.

`:.` Required number

`= 3 xx text()^5P_2 = 3 xx (5!)/(3!) = 60`
Correct Answer is `=>` (C) 60
Q 2339523412

What is the value of `n`, if `P(15, n -1):`
`P (16, n - 2) = 3 : 4 ?`
NDA Paper 1 2011
(A)

10

(B)

12

(C)

14

(D)

15

Solution:

` ∵ (text()^(15)P_( n - 1))/( text()^(16)P_( n - 2)) = 3/4` (given) `[ ∵ text()^nP_r = (n!)/((n - r) !)]`

` => (15 !)/((15-n+1)!) xx ((16 - n + 2)!)/(16!) = 3/4 => ((18-n)!)/(16(16-n)!) = 3/4`

`=> (18- n)(17- n) = 12 => 306 - 17n -18n + n^2 = 12`

`=> n^2 - 35n + 294 = 0 => (n - 14) (n - 21) = 0`

`:. n = 14`

Here `n != 21` because in`text()^nP_r ; n >= r`.
Correct Answer is `=>` (C) 14
Q 2389723617

What is the number of three-digit odd numbers
formed by using the digits `1, 2, 3, 4, 5` and `6`, if
repetition of digits is allowed?
NDA Paper 1 2010
(A)

60

(B)

108

(C)

120

(D)

216

Solution:

Extreme left place can be filled up in `6` ways, the middle

place can be filled up in `6` ways and extreme right place in only

`3` ways.

`:.` Required number of numbers `= 6 xx 6 xx 3 = 108`
Correct Answer is `=>` (B) 108
Q 2309234118

How many words, with or without meaning can
be formed by using all the letters of the word
'MACHINE', so that the vowels occurs only the
odd positions'?
NDA Paper 1 2008
(A)

`1440`

(B)

`720`

(C)

`640`

(D)

`576`

Solution:

There are three vowels and they have four odd places

to arrange. Other letters are four and have four places to arrange.

`:.` The number of words `= text()^4P_3 xx 4 !`

` = (4!)/((4- 3)!) xx 4! = 4! xx 4!`

`= 24 xx 24 = 576`
Correct Answer is `=>` (D) `576`

Circular permutation

(i) The number of circular permutations of n different things taken all at a time is `(n - 1 )!` . If clockwise and
anti-clockwise orders are taken as different.

(ii) If clockwise and anti -clockwise circular
permutations are considered to be the same, then it
is `( (n-1)! )/2`
Q 1542780633

Seven people are seated in a circle. How many relative arrangements are possible?
BITSAT 2013
(A)

`7!`

(B)

`6!`

(C)

`text()^7P_6`

(D)

`text()^7C_6`

Solution:

The number of circular arrangement for n objects is `(n-1)!`.

There would be `(7 - 1)! = 6!` arrangements possible.
Correct Answer is `=>` (B) `6!`
Q 1113080849

The number of ways of arranging `8` men and `4` women around a circular table such that no two women can sit together is
EAMCET 2007
(A)

`8!`

(B)

`4!`

(C)

`8!4!`

(D)

`7!xxtext()^8P_4`

Solution:

Since it is a circular table, the number of arrangements are given by `(n-1)!` therefore all men will arrange themselves by `7`! ways.

For women, no two women can sit together, they will select their seats from `8` places between men.` rArr text()^8P_4`
Hence,

`7!xxtext()^8P_4`
Correct Answer is `=>` (D) `7!xxtext()^8P_4`

Restricted Selection/ Arrangement

1. The number of ways in which r objects can be
selected from n different objects, if k particular
objects are

(a) always included `= text()^(n-k)C_(r-k)`

(b) never include `=text()^(n-k)C_r`


2. The number of arrangements of n distinct objects
taken rat a time, so that!? particular objects are

(a) always included= `text()^(n-k)C_(r-k) * r!`

(b) never include `= text()^(n-k)C_(r ) * r!`
Q 2258823704

The number of ways in which `20` mangoes may be distributed among `5` children, so

that each child gets at least one mango is
BITSAT Mock
(A)

`text()^(20)C_4`

(B)

`text()^(19)C_4`

(C)

`text()^(24)C_3`

(D)

`text()^(19)C_5`

Solution:

The number of ways = The number of positive integral solutions of

`x + y + z + t + u = 20 = text()^(19)C_4` (by partition method).
Correct Answer is `=>` (B) `text()^(19)C_4`
Q 1775380266

In how many ways can a football team of `11` players be selected from `16`
players? How many of them will
(i) include `2` particular players?
(ii) exclude `2` particular players?
NCERT Exemplar
Solution:

Total number of players `= 16`

We have to select a team of `11` players

(i) include `2` particular players `= text()^(16 - 2)C_(11 -2) = text()^(14)C_9`

[since, selection of `n` objects taken `r` at `a` time in which `p` objects are always

included is `text()^(n- p)C_(r- p) ]`

(ii) Exclude `2` particular players `= text()^(16-2)C_(11) = text()^(14)C_(11)`

[since, selection of `n` objects taken `r` at `a` time in which `p` objects are never included

is ` text()^(n-p)C_r]`

Applications : Word formation and Geometrical

Q 2211212129

A polygon has `44` diagonals. The number of its sides is
NDA Paper 1 2015
(A)

`11`

(B)

`10`

(C)

`8`

(D)

`7`

Solution:

Let the number of sides of the polygon be `n`.

Then, number of diagonals `= text()^(n)C_2 - n`

`=> 44 =text()^(n)C_2 - n`

` => 44 = (n!)/(2!(n-2)!) - n `

`=> 44 = ( n (n-1) (n-2)!)/(2!(n-2)!) -n`

` => 44 = [ n (n-1)]/2 -n => 88 = n^2 -n -2n`

`n^2 -3n-88=0`

`=> (n- 11) (n+8)= 0 => n= 11 ,-8`

Hence, polygon has `11` sides.
Correct Answer is `=>` (A) `11`
Q 1743178043

How many different words can be formed by taking four
letters out of the letters of the word `AGAIN`, if each
word has to start with `A`?
NDA Paper 1 2014
(A)

`6`

(B)

`12`

(C)

`24`

(D)

None of these

Solution:

A .. ... .. Fixed.

The word 'AGAIN' has five letters `2A, 1 G, 1I, 1 N`. Since,` A`

repeat two times and `A` is fixed at first position then, we

have to arrange remaining `4` letters in three vacant position.

`:.` Required number of ways `= 1 xx 4 xx 3 xx 2 = 24`
Correct Answer is `=>` (C) `24`
Q 2369212115

In how many ways can the letters of the word
'GLOOMY' be arranged so that the two O's
should not be together?
NDA Paper 1 2013
(A)

`240`

(B)

`480`

(C)

`600`

(D)

`720`

Solution:

First we arrange the four letters G,L,M,Y in the alternate

position `= 4!`

`G, L, M, Y`

Now, rest of letters `0,0` arrange in `5` alternate positions `= 5C_2`

`:.` Required number of ways `= 4! xx text()^5C_2`

` = 24 xx (5 xx 4)/2`

`= 24 xx 10 = 240`
Correct Answer is `=>` (A) `240`
Q 2359412314

What is the number of diagonals which can be
drawn by joining the angular points of a polygon
of `100` sides?
NDA Paper 1 2012
(A)

`4850`

(B)

`4950`

(C)

`5000`

(D)

`10000`

Solution:

The number of diagonals which can be drawn by joining

the angular points of a polygon of `100` sides `= text()^(100)C_2 - 100`

` = (100!) /(2! 98!) - 100`

`= (100 xx 99 xx 98!)/(2 xx 98!) - 100`

`= 50 xx 99 - 100`

` = 4950 - 100`

`= 4850`
Correct Answer is `=>` (A) `4850`
Q 2349623513

What is the maximum number of straight lines
that can be drawn with any four points in a plane
such that each line contains at least two of these
points?
NDA Paper 1 2010
(A)

2

(B)

4

(C)

6

(D)

12

Solution:

Required number of lines `= text()^4C_2 = (4!)/(2! 2!) = 6`
Correct Answer is `=>` (C) 6
Q 2309734618

If `7` points out of `12` are in the same straight line,
what is the number of triangles formed?
NDA Paper 1 2008
(A)

`84`

(B)

`175`

(C)

`185`

(D)

`201`

Solution:

Required number of triangles formed

`= text ()^(12)C_3 - text()^7C_3`

` = (12!)/(3!9!) - (7!)/(3!4!) ( ∵ text()^nC_r = (n!)/(r!( n - r)!))`

` = (12.11.10)/(3.2.1) - (7·6·5)/(3·2·1)`

`= 220 - 35 = 185`
Correct Answer is `=>` (C) `185`
Q 2580191917

Suppose p points are chosen on each of the three coplanar lines. The maximum number of triangles formed with vertices at these points is
BCECE Stage 1 2014
(A)

`p^3 + 3p^2`

(B)

`1/2(p^3 + p)`

(C)

`p^2/2 (5p -3)`

(D)

`p^2 (4 p -3)`

Solution:

Total number of points in a plane is 3p.

`:.` Maximum number of triangles

`=text()^(3p) C_3 - 3 * text()^p C_3`

`= ((3p)!)/((3p -3)! 3!) - 3 (p!)/((p-3) 3!)`

`=(3p (3p -1) (3p -2))/(3 xx 2) - (3 xx p (p -1)(p -2))/( 3xx 2)`

`= p/2 [ 9p^2 -9p + 2 - (p^2 - 3p + 2)]`

`=p^2 (4p -3)`
Correct Answer is `=>` (D) `p^2 (4 p -3)`
Q 2572191936

The number of diagonals in a polygon is `20`.
The number of sides of the polygon is
WBJEE 2011
(A)

`5`

(B)

`6`

(C)

`8`

(D)

`10`

Solution:

Let `n` be a side of polygon.

Then, number of diagonals of the polygon


`= text()^(n)C_2 - n`

`=> 20= (n (n-1) )/2 -n`

`=> n^2 -3n -40 =0`

`=> (n+5)(n-8) =0`

`=> n=8` ( `·: n` cannot be negative)
Correct Answer is `=>` (C) `8`
Q 2315691569

The sum of all the numbers that can be
formed with all of the digits `2, 3, 4, 5` is
BITSAT Mock
(A)

`94324`

(B)

`93324`

(C)

`92324`

(D)

`95324`

Solution:

Total number of digits that can be
formed with `2, 3, 4, 5 = ∟4 = 24` .

`2 + 3 + 4 + 5 = 14`.

Each of the four numbers occur in the unit place `6` times, in the `10^(th)`
place `6` times etc.

`:.` the required sum

`= 6 xx 14 (1 + 10 + 100 + 1000)`

`= 84 xx 1111 = 93324`
Correct Answer is `=>` (B) `93324`
 
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