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previous year questions of Permutation and Combination for NDA

previous year questions of Permutation and Combination

previous year questions
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 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 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 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 2167812785

What is the number of different messages that can be
represented by three ` 0's` and two `1's?`
NDA Paper 1 2016
(A)

`10`

(B)

`9`

(C)

`8`

(D)

`7`

Solution:

Here,we are given three 0's and two 1's.

Hence, number of ways of different messages

` = (5!)/(3! xx 2!) = (5 xx 4)/2 = 10`
Correct Answer is `=>` (A) `10`
Q 1618523400

Let `A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.` Then, the
number of subsets of `A` containing exactly two
elements is
NDA Paper 1 2015
(A)

`20`

(B)

`40`

(C)

`45`

(D)

`90`

Solution:

Required number of subsets of `A` containing

exactly two elements

` = text()^(10)C_2 = (10 xx 9)/2 = (90)/2 = 45`
Correct Answer is `=>` (C) `45`
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 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 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 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 2211612529

The number of ways in which `3` holiday tickets can be
given to `20` employees of an organisation, if each
employee is eligible for any one or more of the tickets, is
NDA Paper 1 2015
(A)

1140

(B)

3420

(C)

6840

(D)

8000

Solution:

Since, each employees is eligible for one or more

ticket(s).

Hence, total number of ways `= 20^3 = 8000`
Correct Answer is `=>` (D) 8000
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 1712478330

`C(n, r): C(n, r + 1) = 1:2` and `C(n, r + 1): C(n,r + 2) =2: 3`.

What is `n` equal to?
NDA Paper 1 2014
(A)

`11`

(B)

`12`

(C)

`13`

(D)

`14`

Solution:

Given that, `C (n, r): C (n. r + 1) = 1:2`

`=> text()^(n) C_r :text()^( n)c_(r-1) + 1 = 1 : 2`

` => ((n!))/(r! (n- r)!)/((n!))/( (r + 1)! (n - r - 1)!) = 1/2`

` => ( (r + 1)! (n - r - 1)!)/ (r!(n-r)!) = 1/2`

` => ( (r + 1).r ! (n - r - 1)!)/( r!(n -r)!(n-r-1)!) = 1/2`

` => (r + 1)/(n - r) = 1/2`

` => 2r + 2 = n -r`

` => n = 3r + 2 ` ...........(1)

and ` C (n, r + 1): C (n, r + 2) = 2 : 3`

` => text()^(n) C_(r+1) : text()^(n) C_(r+2) = 2 : 3`

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

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

` =>((r+2).(r+1)!(n-r-2)!) /( (r + 1)! (n - r -1)(n - r - 2)!) = 2/3`

` => ( r+2)/(n-r-1) = 2/3 `

` => 3r + 6=2n -2r -2`

` => 5r + 8 = 2n ` .........(2)

Now, put the value of r in Eq. (1), we get

` n =3(4)+2`

`n = 12 + 2`

`n = 14`
Correct Answer is `=>` (D) `14`
Q 1732478332

`C(n, r): C(n, r + 1) = 1:2` and `C(n, r + 1): C(n,r + 2) =2: 3`.

What is `r` equal to?
NDA Paper 1 2014
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Given that, `C (n, r): C (n. r + 1) = 1:2`

`=> text()^(n) C_r :text()^( n)c_(r-1) + 1 = 1 : 2`

` => ((n!))/(r! (n- r)!)/((n!))/( (r + 1)! (n - r - 1)!) = 1/2`

` => ( (r + 1)! (n - r - 1)!)/ (r!(n-r)!) = 1/2`

` => ( (r + 1).r ! (n - r - 1)!)/( r!(n -r)!(n-r-1)!) = 1/2`

` => (r + 1)/(n - r) = 1/2`

` => 2r + 2 = n -r`

` => n = 3r + 2 ` ...........(1)

and ` C (n, r + 1): C (n, r + 2) = 2 : 3`

` => text()^(n) C_(r+1) : text()^(n) C_(r+2) = 2 : 3`

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

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

` =>((r+2).(r+1)!(n-r-2)!) /( (r + 1)! (n - r -1)(n - r - 2)!) = 2/3`

` => ( r+2)/(n-r-1) = 2/3 `

` => 3r + 6=2n -2r -2`

` => 5r + 8 = 2n ` .........(2)

Put the value of n from Eq. (1) in Eq. (2), we get

`5r+8=2(3r+2)`

`=> 5r + 8= 6r + 4`

` => 6r- 5r = 8-4`

`:. r = 4`
Correct Answer is `=>` (C) `4`
Q 1702478338

`C(n, r): C(n, r + 1) = 1:2` and `C(n, r + 1): C(n,r + 2) =2: 3`.

What is `P(n, r): C(n, r)` equal to?
NDA Paper 1 2014
(A)

`6`

(B)

`24`

(C)

`120`

(D)

`720`

Solution:

Given that, `C (n, r): C (n. r + 1) = 1:2`

`=> text()^(n) C_r :text()^( n)c_(r-1) = 1 : 2`

` => ((n!))/(r! (n- r)!)/((n!))/( (r + 1)! (n - r - 1)!) = 1/2`

` => ( (r + 1)! (n - r - 1)!)/ (r!(n-r)!) = 1/2`

` => ( (r + 1).r ! (n - r - 1)!)/( r!(n -r)!(n-r-1)!) = 1/2`

` => (r + 1)/(n - r) = 1/2`

` => 2r + 2 = n -r`

` => n = 3r + 2 ` ...........(1)

and ` C (n, r + 1): C (n, r + 2) = 2 : 3`

` => text()^(n) C_(r+1) : text()^(n) C_(r+2) = 2 : 3`

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

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

` =>((r+2).(r+1)!(n-r-2)!) /( (r + 1)! (n - r -1)(n - r - 2)!) = 2/3`

` => ( r+2)/(n-r-1) = 2/3 `

` => 3r + 6=2n -2r -2`

` => 5r + 8 = 2n ` .........(2)

We have, `P (n,r): C (n,r) = text()^(n)p_r : text()^(n)c_r`,

` = (n!) /((n-r)!) : (n!)/(r!(n-r)!)`

` = 1 : = 1/(r!) = r! : 1`

` = 4! : 1 = 24 : 1 `

` = 24`
Correct Answer is `=>` (B) `24`
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 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 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 2339112012

If `C(28, 2 r) = C(28, 2r - 4)`, then what is `r` equal to?
NDA Paper 1 2013
(A)

`7`

(B)

`8`

(C)

`12`

(D)

`16`

Solution:

Given, `C(28, 2r) = C(28, 2r - 4)`

`=> text()^(28)C_(2r) = text()^(28)C_(2r - 4) ( ∵ text()^nC_x = text()^nC_y => x + y = n)`

`=> 2r + (2r - 4) = 28 => 4r = 32`

`:. r = 8`
Correct Answer is `=>` (B) `8`
Q 2319212110

If `P (77, 31) = x` and `C (77, 31) = y`, then which one
of the following is correct ?
NDA Paper 1 2013
(A)

`x = y`

(B)

`2x = y`

(C)

`77x = 31y`

(D)

`x > y`

Solution:

Given that, `P (77, 31) = x`

i.e., `text()^(77)P_(31) = x` .....(i)

and `C (77, 31) = y`

i.e., `text()^(77)C_(31) = y`

From Eq. (ii),

` (77!)/(31! (77! - 31)!) = y => (77!)/((77- 31)!) = 31!y`

` => text()^(77)P_(31) = 31!y [ ∵ text()^nP_r = (n!)/((n - r)!) ]`

` => x = (31!y)` [from Eq. (i))

` :. x > y`
Correct Answer is `=>` (D) `x > y`
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 2389212117

Two numbers are successively drawn from the set
`U = {1,2,3,4,5,6, 7,8}`, the second being drawn
without replacing the first. The number of
elementary events in the sample is
NDA Paper 1 2013
(A)

`64`

(B)

`56`

(C)

`32`

(D)

`14`

Solution:

`:.` Required number of elementary events in the sample

`= text()^8C_2 xx 2! = (8 xx 7)/2 xx 2 = 56`
Correct Answer is `=>` (B) `56`
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 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 2309412318

What is the number of ways that `4` boys and `3`
girls can be seated so that boys and girls alternate?
NDA Paper 1 2012
(A)

`12`

(B)

`72`

(C)

`120`

(D)

`144`

Solution:

The required number of ways that `4` boys and `3` girls

can be seated, so that boys and girls alternate `= 4! xx 3!`

`= 24 xx 6 = 144`
Correct Answer is `=>` (D) `144`
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 2359612514

What is the value of ` sum_(r = 1)^n (P(n, r))/(r!)` ?
NDA Paper 1 2011
(A)

`2^n - 1`

(B)

`2^n`

(C)

`2^(n- 1)`

(D)

`2^n + 1`

Solution:

` sum_(r = 1)^n (P(n, r))/(r!)`

` = sum_(r = 1)^n 1/(r!) . (n!)/(( n - r)!) quad ( ∵ text()^nP_r = (n!)/(( n - r)!))`

` = sum_(r = 1)^n 1/(r!) text()^nC_r quad ( ∵ text()^nC_r = (n!)/(r!( n - r)!))`

`= ( text()^nC_1 + text()^nC_2 + text()^nC_3 + ··· + text()^nC_n)`

`= (1 + text()^nC_1 + text()^nC_2 + text()^nC_3 + ··· + text()^nC_n) -1`

`= ( text()^nC_0 + text()^nC_1 + text()^nC_2 + ··· + text()^nC_n) - 1`

`= (1 + 1)^n - 1 = 2^n - 1`
Correct Answer is `=>` (A) `2^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 2379223116

What is the total number of combinations of `n`
different things taken `1, 2, 3, ... ,n` at a time?
NDA Paper 1 2011
(A)

`2^(n+1)`

(B)

`2^(2n+1)`

(C)

`2^(n-1)`

(D)

`2^n - 1`

Solution:

Since, combinations formed after taking `1, 2, 3, ... , n`

things at a time are ` text()^nC_1, text()^nC_2 ... , text()^nC_n`.

`:. ` Total number of combinations `= text()^nC_1 + text()^nC_2 + ... + text()^nC_n`

`= 1 + text()^nC_1 + text()^nC_2 + ... + text()^nC_n - 1`

`= 2^n - 1 ( ∵ 2^n = text()^nC_0 + text()^nC_1 + text()^nC_2 + ... + text()^nC_n`)
Correct Answer is `=>` (D) `2^n - 1`
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 2349323213

In how many ways `6` girls can be seated in two
chairs?
NDA Paper 1 2011
(A)

10

(B)

15

(C)

24

(D)

30

Solution:

Required number of ways `= 6 xx 5 = 30`
Correct Answer is `=>` (D) 30
Q 2389323217

If `x + y <= 4`, then there are how many non-zero
positive integer ordered pair `(x, y)`?
NDA Paper 1 2011
(A)

4

(B)

5

(C)

6

(D)

8

Solution:

The possible non-zero positive integer ordered pair

`(x, y)` satisfy the inequality `x + y <= 4` is `(1, 1),(1, 2),(1, 3),(2, 1),(2,2)`,

`(3, 1)`.

`:.` Number of required ordered pairs `= text()^4C_2 = ( 4 xx 3)/2 = 6`
Correct Answer is `=>` (C) 6
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 2349523413

Using the digits `1, 2, 3, 4` and `5` only once, how
many numbers greater than `41000` can be formed?
NDA Paper 1 2011
(A)

41

(B)

48

(C)

50

(D)

55

Solution:

Required number of ways `= 2 xx 4 xx 3 xx 2 xx 1 = 48.`
Correct Answer is `=>` (B) 48
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 2319523419

What is the number of words that can be formed
from the letters of the word 'UNIVERSAL', the
vowels remaining always together?
NDA Paper 1 2010
(A)

720

(B)

1440

(C)

17280

(D)

21540

Solution:

Since, the vowels in the word 'UNIVERSAL' are U, I, E

and A.

Let us consider these as a single letter.

Then, number of ways to arrange them `= 6! = 720`

But vowels can also arranged in `4 !` or `24` ways.

Hence, total number of ways `= 720 xx 24 = 17280`
Correct Answer is `=>` (C) 17280
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 2389623517

What is the number of signals that can be sent by
`6` flags of different colours taking one or more at a
time?
NDA Paper 1 2010
(A)

21

(B)

63

(C)

720

(D)

1956

Solution:

Required number of ways `= 2^6 - 1 = 64 - 1 = 63 .`
Correct Answer is `=>` (B) 63
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 2369723615

What is the number of ways of arranging the
letters of the word 'BANANA' so that no two `N`'s
appear together?
NDA Paper 1 2010
(A)

`40`

(B)

`60`

(C)

`80`

(D)

`100`

Solution:

Number of ways that can be formed by using the

words 'BANANA' `= (6!)/(3!2!) = 60`

Number of ways in which two `N` comes together `= (5!)/(3!) = 20`

`:.` Required number of ways `= 60 - 20 = 40`
Correct Answer is `=>` (A) `40`
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 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 2359823714

In a football championship `153` matches were
played. Every team played one match with each
other team. How many teams participated in the
championship?
NDA Paper 1 2009
(A)

`21`

(B)

`18`

(C)

`17`

(D)

`15`

Solution:

Let total number of teams that participated in a championship be `n`.

` :. text()^nC_2 = 153 => (n(n - 1))/2 = 153`

` => n(n -1) = 306`

`:. n = 18`
Correct Answer is `=>` (B) `18`
Q 2369023815

A coin is tossed `10` times. The number of heads
minus the number of tails in `10` tosses is
considered as the outcome of the experiment.
What is the number of points in the sample
space?
NDA Paper 1 2009
(A)

`10`

(B)

`11`

(C)

`21`

(D)

`99`

Solution:

Hence, total number of points in the sample space is `11`.
Correct Answer is `=>` (B) `11`
Q 2309023818

Two numbers are successively drawn from the set
`{ 1, 2, 3, 4, 5, 6, 7}` without replacement and the
outcomes recorded in that order. What is the
number of elementary events in the random
experiment?
NDA Paper 1 2009
(A)

`49`

(B)

`42`

(C)

`21`

(D)

`14`

Solution:

Total number of elementary events `= text()^7C_1 xx text()^6C_1 = 42`
Correct Answer is `=>` (B) `42`
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`
 
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