Mathematics Must Do Problems of Permutation and Combination for NDA

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Must Do Problems of Permutation and Combination for NDA

Must Do Problems of Permutation and Combination

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)

Q 1382856737

A person wishes to make as many different parties as he can out of `10` of his friends. Each party consist of same no. of friends. Out of these parties in how many parties a particular friend will be found?

(A)

`252`

(B)

`10`

(C)

`210`

(D)

`126`

Solution:

Let him invite r friends at a time in each party.

Then the number of parties `= text()^{10}C_r`

This is maximum only if `r=5`

Out of these `text()^{10}C_5` parties a particular person can be found in `text()^9C_4 =126` parties.

This is because out of any `5` friends `1` particular friend is included and the remaining `4` friends are to be selected from remaining `9` friends. `=text()^9C_4` (ways.)

Correct Answer is `=>` (D) `126`

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 2353012844

In a football championship, `120` matches were played. Every two teams played
one match with each other. Then the number of participant teams are
BITSAT Mock

(A)

`18`

(B)

`15`

(C)

`16`

(D)

`17`

Solution:

If `n` is the number of teams, it is
given by

`text()^(n)C_2 = 120`

`=> (n (n-1) )/2 =120`

`=> n^2 -n -240 = 0`

`=> (n+15) (n-16) =0`

`=> n =16`

Correct Answer is `=>` (C) `16`

Q 1510745619

Total number of books is `2n + 1`. One is allowed to select a minimum of the one book and a maximum of `n` books. If total number of selections if `63`, then value of `n` is :
BITSAT 2005

(A)

`3`

(B)

`6`

(C)

`2`

(D)

None of these

Solution:

Since
`(1 + x)^(2n- 1) = C_0 + C_1 x + ... + C_n x^n + C_(n+1) x^(n+1) + ........+ x^(2n+1)`

`= 2(C_0 +C_1 +...............+ C_nx^n)`

Put `x= 1`

`=> (1 + 1)^(2n-1) =2(C_0 +C_1 + ... +C_n)`

`=> 2^(2n) =(C_1 +C_1 +.....C_n)`

`=> 2^(2n) -1 =C_1 +C_1 +.....C_n`

`=> 2^(2n) -1 =63`

`=> 2^(2n) = 64 => 2^(2n) =2^6`

`=> 2n =6 => n =3`

Correct Answer is `=>` (A) `3`

Q 2834591452

If `text()^(2n + 1) P_(n -1) : text ()^(2n - 1 ) P_n = 3 : 5`, then consider the following statements

I. The value of n is 3
II. P(5, n) = P(6, n -1)
Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We have, `(text()^(2n +1 ) P _(n -1))/(text()^(2n -1) P_n) = 3/5`

`=> ((2n +1) !)/((n +2 ) ! ) xx ((n -1 ) !)/( ( 2 n -1 )!) = 3/5`

`=> ((2n+1)(2n)(2n-1)!)/((n+2)(n+1)n(n-1)!) xx ((n-1)!)/((2n -1)!) =3/5`

`=> (2( 2n +1))/( ( n +2) ( n +1) = 3/5`

`=> 10(2n+ 1) = 3(n+ 2)(n+ 1)`

`=> 3n^2 + 9n + 6 = 20n + 10`

`=> 3n^2 -11 n - 4 = 0`

`=> (n- 4) (3n + 1) = 0 => n = 4`

`:.` Statement I is not correct.

Now , for `n = 4`

`P( 5, n ) = text()^5P_4 = (5 !)/( 1 !) = 120`

and `P( 6 , n -1) = text()^6 _3 = (6 !)/( 3 !) = 120`

`=> P(5, n) = P ( 6 , n -1)` for `n =4`

`:. ` statement II is correct

Correct Answer is `=>` (B) Only II

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)

(B)

(C)

(D)

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 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`

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]`

Q 2536178072

In how many ways can `9` different books be distributed among three students if each receives atleast `2` books?

Solution:

If each receives atleast `2` books, then the division as shown by tree diagrams

The number of division ways for tree diagrams (i), (ii) and (iii) are

` (9!)/((2 !)^2 (5 !)) xx 1/(2!) , (9!)/(2! 3 !4!)` and `(9!)/(3!)^3 xx 1/(3!) `, respectively.

Hence, the total number of ways of distribution of these groups among

`3` students is

`[ (9!)/( (2 !)^2 (5 !)) xx 1/(2!) + (9!)/( 2! 3! 4!) + (9!)/(3 !)^3 xx 1/(3!) ] xx 3!`

`= [ 378 + 1260 + 280 ] xx 6 = 11508`

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 2562012835

How many triangles can be formed by joining `6` points lying on a circle?
WBJEE 2011

Solution:

Number of such triangle `= text()^(6)C_3`

`= (6!)/(3! 3!)= 20`

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`

Correct Answer is `=>` (C) `8`

Q 2874391256

A parallelogram is cut by two sets of m para11el to its sides. The number parallelograms thus formed is

(A)

`text()^mC_2 xx text()^mC_2`

(B)

`2(text()^(m +2)C_2)`

(C)

`(text()^(m +2) C_2)^2`

(D)

None of these

Solution:

The two sets of m parallel lines along with two sets of two parallel lines of the given parallelogram will form two sets of (m + 2) parallel lines. Each parallelogram is formed by choosing two parallel lines from each of the above.

`:.` Total number of parallelograms

`= text()^( m +2 ) C_2 xx text()^(m +2 ) C_2 = (text()^( m +2 ) C_2 )^2`

Correct Answer is `=>` (C) `(text()^(m +2) C_2)^2`

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`

Q 2364278155

The sum of the digits in the unit place
of all numbers formed with the help of
`3, 4, 5, 6` taken all at a time is :
BITSAT Mock

(A)

`144`

(B)

`432`

(C)

`108`

(D)

`18`

Solution:

When number at unit place is `3`, then

other `3` numbers can be arranged in `3 !`

ways.

`∴` The sum of the digits in unit place

when `3` is their at unit place `= 3 ! × 3` ...(1)

Similarly,

the sum of the digits in .... when

`4` is `.... = 3 ! × 4` ...(2)

`5 .... = 3 ! × 5` ...(3)

`6 .... = 3 ! × 6` ...(4)

from (1), (2), (3), (4)

`∴` The sum of the digits in the unit place

of all numbers formed with the help of

`3, 4, 5, 6` taken all at a time is

`(3 ! × 3) + (3 ! × 4) + (3 ! × 5) + (3 ! × 6)`

`= 3 ! (3 + 4 + 5 + 6)`

`= 6 × 18`

`= 108`.

Correct Answer is `=>` (C) `108`

Set 2

Q 2514156050

How many `10` digits numbers can be written by
using digits (`9` and `2` )?
UPSEE 2008

(A)

` text()^(10)C_1 + text()^(9)C_2`

(B)

`2^(10)`

(C)

`text()^(10)C_2`

(D)

`10!`

Solution:

Total numbers of numbers `= 2 xx 2 xx ... xx 10` times

`= 2^(10)`

Correct Answer is `=>` (B) `2^(10)`

Q 2511167920

Two decks of playing cards are well shuffled and `26` cards are randomly distributed to a player. Then, the probability that the player gets all distinct cards is
WBJEE 2012

(A)

` text()^(52)C_(26) // text()^(104)C_(26)`

(B)

` 2 xx text()^(52)C_(26) // text()^(104)C_(26)`

(C)

`2^(13) xx text()^(52)C_(26) // text()^(104)C_(26)`

(D)

`2^(26) xx text()^(52)C_(26) // text()^(104)C_(26)`

Solution:

Since. these are `52` distinct cards in decks and

each distinct card is `2` in number.

Therefore. 2 decks will also contain only `52`

distinct cards two each.

`:.` Probability that the player gets all distinct

cards

` = ( text()^(52)C_(26) xx 2^(26))/(text()^(104)C_(26))`

Correct Answer is `=>` (D) `2^(26) xx text()^(52)C_(26) // text()^(104)C_(26)`

Q 2875401366

(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

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 2874691556

Four-letter words are to be formed using the letters of the word 'FAILURE'.

Consider the following statements
I. Number of words ofF is included in each word is `text()^6C_3 xx 4!`.
II. Number of words, if it contains two different vowels and two different consonants is
`text()^3C_2 xx text()^4C_2 xx 4 !`

Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

There arc 7 letters in the word 'FAILURE'.
Number of vowels = 4
Number of consonants = 3
Out of four letters, one letter is F and other three letters can be selected from remaining 6 letters in `text()^6 C_3` ways.

These four letters can be 'arranged in 4! ways.

Total number of words `= text()^6 C_3 xx 4 !`

So, Statement I is correct.

Now, 2 vowels and 2 consonants can be chosen in `text()^3C_2 xx text()^4C_2` ways and can be arranged in `4!` ways

Total number of words

`= text()^3C_2 xx text()^4C_2 xx 4 !`

So, Statement II is correct.

Correct Answer is `=>` (C) Both I and II

Q 2560334215

The number of six-digit numbers that can be
formed from the digits 1, 2, 3, 4, 5, 6 and 7, so
that digits do not repeat and the terminal
digits are even is
BCECE Stage 1 2013

(A)

`144`

(B)

`72`

(C)

`288`

(D)

`720`

Solution:

Terminal digits are the first and last digits.

Since, terminal digits are even.

`:. 1`st place can be filled in `3` ways and last

place can be filled in `2` ways and remaining

places can be filled in

`text()^5P_4 = 120` ways

Hence, the number of six digit number, the

terminal digits are even, is `3 xx 120 xx 2 = 720`.

Correct Answer is `=>` (D) `720`

Q 2449678513

The number of ways in which 5 boys and 5 girls can be seated for a photograph, so that no two girls sit next to each other is
BCECE Stage 1 2012

(A)

`6!5!`

(B)

`(5!)^2`

(C)

`(10!)/(5!)`

(D)

`(10!)/(5!)^2`

Solution:

Required number of ways = 5! x 6!

Correct Answer is `=>` (A) `6!5!`

Q 2572691536

Out of 7 consonants and 4 vowels, the number of words (not necessarily meaningful) that can be made, each consisting of 3 consonants and 2 vowels, is
WBJEE 2014

(A)

`24800`

(B)

`25100`

(C)

`25200`

(D)

`25400`

Solution:

3 consonants can be selected from 7 consonants

`= text()^7 C_3` ways

2 vowels can be selected from 4 vowels

`=text()^4 C_2` ways

`:.` Required number of words

`=text()^7 C_3 xx text()^4 C_2xx 5!`

[selected 5 letters can be arrange in `5!`, so get, a different words]

`=35 xx6 xx120=25200`

Correct Answer is `=>` (C) `25200`

Q 2875801766

Different words are being formed by arranging the letters of the word
'ARRANGE'
The number of arrangement in which two R's are never together is

(A)

900

(B)

1080

(C)

1020

(D)

960

Solution:

There arc 7 letters in the word 'ARRANGE' and there are two Ns and two R's and three different letters.

Number of consonants = 4

Number of vowels = 3
Total number of arrangement when there is no restriction

`= (7! )/( 2 ! 2!) = 1260`

Number of arrangement, when two R's are together

`= ( 6! 2! ) / ( 2 ! 2 ! ) =360`

`:.` Required number ` =1260 -360 = 900`

Correct Answer is `=>` (A) 900

Q 1582780637

In how many ways can `4` people be seated on a square table, one on each side?
BITSAT 2013

(A)

`4!`

(B)

`3!`

(C)

`1`

(D)

None of these

Solution:

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

Similar to arrangements in a circle, there would be `3! ` ways possible of making `4` people sit on a square table.

Correct Answer is `=>` (B) `3!`

Q 1312434339

The no. of ways in which `20` different pearls of two colours can be set alternatively on a necklace, there being `10` pearls of each colour, is ?

(A)

`5*(10!)^2`

(B)

`5*(9!)^2`

(C)

`(9!)^2`

(D)

`9!*10!`

Solution:

Pearls of one colour can be arranged in `{(10-1)!}/{2}={9!}/{2}`

`10` pearls of the other colour are arranged in `10` places between the pearls of `1^{st}` colour.

in `10!` ways.
`therefore` the required no. of arrangements `= {9!}/{2} times 10!=5 times (9!)^2`

Correct Answer is `=>` (B) `5*(9!)^2`

Q 2539578412

The number of ways of distributing `8` identical balls in `3` distinct boxes, so that None of the boxes is empty, is
BCECE Mains 2015

(A)

`5`

(B)

`21`

(C)

`text()^8C_3`

(D)

`3^8`

Solution:

The total number of ways of dividing `8` identical

balls in `3` distinct boxes so that none of the boxes is

empty ` = text()^(8 - 1)C_(3 - 1) = text()^7C_2`

` = (7!)/(5!2!) = 21`

Correct Answer is `=>` (B) `21`

Q 1716001879

The number of ways in which a team of eleven players can be selected
from `22` players always including `2` of them and excluding `4` of them is
NCERT Exemplar

(A)

` text()^(16)C_(11)`

(B)

` text()^(16)C_(5)`

(C)

` text()^(16)C_(9)`

(D)

` text()^(20)C_(9)`

Solution:

Total number of players `= 22`

We have to select a team of `11` players. Selection of `11` players when `2` of them. is

always included and `4` are never included.

Total number of players `= 22 - 2 - 4 = 16`

`:.` Required number of selections `= text()^(16)C_9`

Correct Answer is `=>` (C) ` text()^(16)C_(9)`

Q 2424391251

The number of triangles that can be formed by
choosing the vertices from a set of `12` points,
seven of which lie on the same straight line, is
UPSEE 2012

(A)

`185`

(B)

`175`

(C)

`115`

(D)

`105`

Solution:

Required number of ways `= text()^(12)C_3 - text()^(7)C_3`

`=220- 35`

`=185`

Correct Answer is `=>` (A) `185`

Q 2541145923

The number of diagonals that can be drawn in a polygon of 15 sides, is :
BCECE Stage 1 2015

(A)

(B)

(C)

(D)

Solution:

If there are insides of a polygon, then

total number of diagonals are `text()^nC _2 - n`.

If there are 15 sides of a polygon, then total number of diagonals

`= text()^15 C_2 - 15`

`= 105 -15`

`=90`

Correct Answer is `=>` (C) 90

Q 2469278115

The sum of the integers from 1 to 100 which are divisible by 3 and 5, is
BCECE Stage 1 2012

(A)

`2317`

(B)

`2632`

(C)

`315`

(D)

`2489`

Solution:

Sum of the integers which are divided both 3 and 5

`= 15 + 30 + 45 + ... + 90`

` = 6/2(15+90)`

` = 315`

Correct Answer is `=>` (C) `315`

Q 2864791655

Find the total numbers of four digits number that are greater than 3000, that can be formed using the digits 1, 2, 3, 4, 5 and 6.
If repetition is not allowed

(A)

240

(B)

(C)

360

(D)

480

Solution:

Required number of numbers

`= 4 xx 5 xx 4 xx 3 = 240`

Correct Answer is `=>` (A) 240

Q 2363812745

Five digit numbers are formed with `0, 1, 3, 5, 7`. The number of numbers in which
at least one digit is repeated is
BITSAT Mock

(A)

`96`

(B)

`2400`

(C)

`2500`

(D)

`2404`

Solution:

The number of all five digit numbers
that can be formed with `0, 1, 3, 5`,

`7 = 4 * 5^4` `( :. 0` cannot occur in the `10000^(th)` place)

`= 2500`

The number of numbers without any

repetition `= 4 xx 4 xx 3 xx 2 xx 1 = 96`

`:. ` the number of five digit numbers

with at least one digit repeated

`= 2500 - 96 = 2404`

Correct Answer is `=>` (D) `2404`