Solution

JEE MATHS PERMUTATION AND COMBINATION

JEE main - Mathematics

1.
(d) 8
Explanation: Since Soham is on the fourth seat, therefore two girls Manu and Tanu cannot occupy the fifth seat. The fifth seat
can be occupied by any of the remaining 2 students i.e. in 2 ways and the fourth seat in 1 way (fixed for Soham). The first three
seats can be occupied by Manu and Tanu in these four ways -
(Manu, Tanu, __ , Soham, __ ),
(Tanu, Manu, __ , Soham, __ )
( __ , Manu, Tanu, Soham, __ ),
( __, Tanu, Manu, Soham,__ )
⇒ Number of arrangement that the two girls can sit adjacent to each other = 4

⇒ The required number of ways = 2 × 1 × 4 = 8

2.
(c) 1240
Explanation: Number of ways of selecting a man and a woman for a team from 15 men and 15 women
= 15 × 15 = (15)2
Number of ways of selecting a man and a woman for next team out of the remaining 14 men and 14 women.
= 14 × 14 = (14)2
Similarly for other teams.
Hence required number of ways
= (15)2 + (14)2 +...+ (1)2 = 15×16×31

6
= 1240

3.
(d) 15
Explanation: 15
4.
(c) 7×8!

Explanation: 7×8!

5. (a) 40
Explanation: Total number of arrangements of the letters of the word ‘BANANA'
6!
= 3!2!
= 60
Number of arrangements in which two N’s come together
5!
= 3!
= 20
⇒ The required number of arrangements
= 60 - 20 = 40
6.
(d) n+2C3
Explanation: Let a, b, c be side lengths
a, b, c ∈ {n + 1, n + 2, ..., 2n}
Case I Case II Case III

Type Scalene △ Isosceles △ Equilateral △

Number of △s nC 2 × nC2 nC
3 1

Required number = nC3 + 2(nC2) + nC1

= (nC3 + nC2) + (nC2 + nC1)

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 = (n+1C3 + n+1C2)

= n+2C3

7.
(b) 225
25
Explanation: Given, ∑ r=0
{
50
Cr .
50−r
C25−r } = K
50
C25

25 50! (50−r)!
50
⇒ ∑ ( × ) = K C25
r=0 r!(50−r)! (25−r)!25!

25 50! 25! 50
⇒ ∑r=0 ( × ) = K C25
25!25! r!(25−r)!

[on multiplying 25! in numerator and denominator.]

50 25 50
⇒ C25 , ∑ 25Cr = K C25
r=0

50 50!
[∵ C25 = ]
25!25!

25 25
⇒ K = ∑ 25Cr = 2
r=0

n n n n n
[∵ C0 + C1 + C2 + … . + Cn = 2 ]

⇒ K = 225
8.
(b) 17! × 2!
Explanation: After placing the 3 persons (1 host + 2 persons on either side) and treating them as 1 person, the total number of
persons = (20 - 3 + 1) = 18
These 18 people can be arranged around the table (circular permutation)
= 17! ways
Again keeping the host fixed, the two particular persons can arrange themselves on either side of the host in 2! ways
Hence, the possible number of arrangements
= 17! × 2!
9.
(b) 24
Explanation: Number of ways in which 4 boys can be seated = (4 - 1)! = 3! = 6
Number of chairs of different colours = 4
⇒ Total number of ways = 6 × 4 = 24

10.
(d) 2 × 5!
5!
Explanation: There are five odd and 4 even positions. Five odd positions can be filled with A, E, E, E, O in 3!
ways and four
even positions can filled with D, N, N, L in 4!
ways
2!

⇒ The required number of permutations

=
5!

3!
× = 2 × 5!
4!

2!

11.
(d) 325
Explanation: Different codes can be made by taking one or two or three or four or five characters at a time. The required
number of codes
= 5P1 + 5P2 + 5P3 + 5P4 + 5P5
= 5 + 20 + 60 + 120+ 120
= 325
12.
(c) None of these
Explanation: Since, the first 2 women select the chairs amongst 1 to 4 in 4P2 ways. Now, from the remaining 6 chairs, three

men could be arranged in 6P3

∴ Total number of arrangements = 4P2 × 6P3

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13. (a) 72
Explanation: Required number should be even and should be divisible by 6.
Then, number should also divisible by 2 & 3.
The possible choices for number divisible by 3 are (1, 2, 5, 6, 7) or (1, 2, 3, 5, 7)
Using 1, 2, 5, 6, 7, number of even numbers is
= 4 × 3 × 2 × 1 × 2 = 48
Using 1, 2, 3, 5, 7, number of even numbers is
= 4 × 3 × 2 × 1 × 1 = 24
Required answer is 72.
14.
(d) 756
Explanation: If we take a pair of 2 alike letters, then the remaining 2 letters are distinct.
There are 3 pairs of alike letters (MM, AA, TT).
Number of ways to choose 1 pair = 3C1 = 3
After choosing 1 pair, 7 letters are available to choose to remain 2 unlike letters from
Number of ways to choose 2 unlike letters = 7C2 = 21
In each of these 4 letters, there are 2 the same letters. Number of ways to arrange 4 letters
4
P4
= = 12
2!

⇒ The required number of ways = 3 × 21 × 12

= 756

15. (a) 34
Explanation: Each number between 1000 and 10000 is a 4-digit number. So we have to form 4-digit numbers where every
digit is from 2 or 3 or 5. Thus digit at thousand’s, hundred’s, ten’s and unit’s place can be filled in 3 ways each.
⇒ Number of numbers between 1000 and 10000
=3×3×3×3
= 34
16. (a) 1620
Explanation: 2nd and 4th row block has to be selected.
One row out of 1st, 3rd, and 5th will contain two letters and remaining one one letter.
Selection of one row that contains two letters is done in 3C1 ​ways.

Now two block can be selected from this row in 3C2 ​ways.

From each of remaining two rows, one blocks can be selected in 3C1 ​ways.
Hence, total selections = 3
C1 ⋅
3
C2 ⋅
3
C1 ⋅
3
C1 = 81
6!
Number of ways of filling A, A, A, B, B, B = 81 × 3!3!

= 81 × 20 = 1620
17.
(d) n+4Cr

Explanation: nCr + 4⋅ nCr-1 + 6⋅ nCr-2 + 4⋅ nCr-3 + nCr-4

= (nCr + nCr-1) + 3(nCr-1 + nCr-2) + 3(nCr-2 + nCr-3) + (nCr-3 + nCr-4)

= n+1Cr + 3⋅ n+1Cr-1 + 3⋅ n+1Cr-2 + n+1Cr-3

= (n+1Cr + n+1Cr-1) + 2(n+1Cr-1 + n+1Cr-2) + (n+1Cr-2 + n+1Cr-3)

= n+2Cr + 2⋅ n+2Cr-1 + n+2Cr-2

= (n+2Cr + n+2Cr-1) + (n+2Cr-1 + n+2Cr-2)

= n+3Cr + n+3Cr-1

= n+4Cr

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18. (a) 233
Explanation:
Number of 2 steps Number of ways Reason

0 1

1 11!
= 11C1 There are 10 single steps S and 1 double step D.
10!1!

2 10!
= 10C2 There are 8 single steps S and 2 double steps D.
8!2!

3 9!
= 9C3 There are 6 single steps S and 3 double steps D.
3!6!

4 8!
= 8C4 There are 4 single steps S and 4 double steps D.
4!4!

5 7!
= 7C5 There are 2 single steps S and 5 double steps D.
5!2!

6 1
11!
(10S’s and 1 D are arranged in 10!1!
ways.
Applying theorem 3 of permutations)
Total number of ways
= 1 + 11C1 + 10C2 + 9C3 + 8C4 + 7C5 + 1
= 1 + 11 + 45 + 84 + 70 + 21 + 1 = 233
19.
(c) 17640
Explanation: 5 red balls from 10 red balls can be drawn in 10C5 ways

4 white balls from 8 white balls can be drawn in 8C4 ways

⇒ Total number of ways
= 10C5 × 8C4
10! 8! 10×9×8×7×6 8×7×6×5
= × = ×
5!5! 4!4! 5×4×3×2×1 4×3×2×1

= 252 × 70
= 17640
20. (a) 1206
Explanation: There are 9 letters (2S, 2H, and 5 distinct letters) out of which there are 7 distinct letters. (D, E, V, A, S, H, I)
Case I: All 4 letters are distinct.
Number of 4 - letter words = 7P4 = 840
Case II: 2 alike and 2 distinct letters
We can choose 1 pair of alike letters from 2 pairs (2S and 2H) and 2 distinct letters from the remaining 6 distinct letters (1 of S
or H and 5 distinct)
Number of ways to choose these letters
= 2C1 × 6C2 = 2 × 15 = 30
Again number of ways to arrange 4 letters
= = 12
4!

2!

Number of such words = 30 × 12 = 360

Case III: Two pairs of alike letters
Since, there are only two pairs of alike letters. so number of ways to choose them = 2C2 = 1
These 4 letters can be arranged in 2!2!
4!
ways = 6 ways
Number of such words = 1 × 6 = 6
⇒ Required number of words

= 840 + 360 + 6
= 1206

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21.
(d) 36
Explanation: 36
22.
(c) 3360

Explanation:

Ways = 8!
× 3! = 8×7×6×5×4

2
= 56 × 60 ⇒ 3360
3!3!2!

23.
(d) 582
Explanation: Since B ... → 5! = 120
C ... → 5! = 120
I ... → 5! = 120
L ... → 5! = 120
PB ... → 4! = 24
PC ... → 4! = 24
PL ... → 4! = 24
PI ... → 4! = 24
PUBC ... → 2! = 2
PUBI ... → 2! = 2
PUBLC ... → 1! = 1
PUBLIC ... → = 1
Serial number = 4(120) + 4(24) + 6 = 582
24.
(c) 190
Explanation: Let there are n balls used to form the sides of equilateral triangle. According to the question, we have
+ 99 = (n - 2)2
n(n+1)

⇒ n2 + n + 198 = 2 [n2 - 4n + 4]
⇒ n2 - 9n - 190 = 0
⇒ n2 - 19n + 10n - 190 = 0
⇒ (n - 19) (n + 10) = 0

⇒ n = 19, - 10
⇒ n = 19 [∵ number of balls n > 0]
n(n+1)
Now, number of balls used to form an equilateral triangle is 2
19×20
= = 190
2

25.
(c) 12 ⋅ 7!
Explanation: 12 ⋅ 7!
26. (a) 6
Explanation: We have, 12Pr = 11P6 + 611P5
12P = 12-1P + 6. 12-1P6-1 ...(i)
⇔ r 6

Also, nPr = n-1Pr + r. n-1Pr-1 ...(ii)

Comparing (i) with (ii), we get
r=6
27.
(d) 2378

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 Explanation: 2378
28.
(b) 3, 6
Explanation: 35Cn+7 = 35C4n-2

⇔ n + 7 = 4n - 2 or n + 7 + 4n - 2 = 35 ...[nCp = nCr ⇒ r = p or p + r = n]
⇔ 3n = 9 or 5n = 30
⇔ n = 3 or n = 6
29.
(b) 300
Explanation: Number of girls in the class = 5 and number of boys in the class = 7
Now, total ways of forming a team of 3 boys and 2 girls
7 5
= C C2 = 350
3

But, if two specific boys are in team, then number of ways = 5 5

C1 ⋅ C2 = 50

Required ways, i.e. the ways in which two specific boys are not in the same team = 350 - 50 = 300
30.
(c) (5N)40
Explanation: (5N)40
31.
(c) 485
Explanation: Given, X has 7 friends, 4 of them are ladies and 3 are men while Y has 7 friends, 3 of them are ladies and 4 are
men.
∴ Total number of required ways

3 4 4 3 3 4 4 3 3 4 4 3 3 4 4 3
= C3 × C0 × C0 × C3 + C2 × C1 × C1 × C2 + C1 × C2 × C2 × C1 + C0 × C3 × C3 × C0

= 1 + 144 + 324 + 16 = 485

32.
(c) 60
Explanation: Since, the sum of given digits 0 + 1 + 2 + 5 + 7 + 9 = 24
Let the six-digit number be abcdef and to be divisible by 11, so the difference of sum of odd placed digits and sum of even
placed digits should be either 0 or a multiple of 11 means | (a + c + e) - (b + d + f) should be either 0 or a multiple of 11.
Hence, possible case is a + c + e = 12 = b + d + f (only)
Now, Case I
set {a, c, e} = {0, 5, 7} and set {b, d, f} = {1,2,9}
So, number of 6-digits numbers = (2 × 2!) × (3!) = 24
[∵ a can be selected in ways only either 5 or 7]
Case II
Set {a, c, e} = {1,2, 9} and set {b, d, f} = {0, 5, 7}
So, number of 6-digits numbers = 3! × 3! = 36
So, total number of 6-digits numbers = 24 + 36 = 60
33.
(d) 120
Explanation: The choice are: 5xxx 0 = 4 × 3 × 2 = 24
7xxxx = 4 × 3 × 2 × 2 = 48
9xxxx = 4 × 3 × 2 × 2 = 48
So required number = 24 + 48 + 48 = 120
34.
(c) 333
Explanation: Total number of triangles

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 = C ⋅ C ⋅ C + C ⋅ C + C ⋅ C +
3
1
5
1
6
1
3
2
5
1
3
2
6
1
5
C2 ⋅
3
C1 +
5
C2 ⋅
6
C1 +
6 3
C2 ⋅ C1 +
6
C2 ⋅
5
C1

= 90 + 15 + 18 + 30 + 60 + 45 + 75 = 333

35. (a) 2(26!)2

Explanation:

Black cards in 26 places are arranged in 26! ways and the red cards in 26 gaps are arranged in 26! ways.
The deck may start with a black or a red card.
Required number = 2(26!) (26!)
= 2(26!)2
36.
(b) 10
Explanation: 10
37.
(b) 12
Explanation: Numbers formed from 3, 4, 5, 6 and which are divisible by 5 have ‘5’ fixed in unit’s place
3 Digit Numbers
HTU
xx5
3P ways
2
3!
= 1!
=3×2
4 Digit Numbers
Th H T U
xxx5
3P ways
3
= 3!

0!
=3×2
⇒ Total number of numbers = 6 + 6 = 12
38.
(d) 220
Explanation: Given that, out of 31 objects 10 are identical and remaining 21 are distinct, so in following ways, we can choose
10 objects
0 identical + 10 distincts, number of ways = 1 × 21C10

1 identical + 9 distincts, number of ways = 1 × 21C9

2 identicals + 8 distincts, number of ways = 1 × 21C8

So, total number of ways in which we can choose 10 objects is
21C + 21C9 + 21C8 + ... + 21C0 = x(let) ...(i)
10
21C + 21C + 21C13 + ... + 21C21 = x ...(ii) [∵nCr = nCn-r]
⇒ 11 12
On adding both Eqs. (i) and (ii), we get
2x = 21C0 + 21C1 + 21C2 + ... + 21C10 + 21C11 + 21C12 + ... + 21C21

⇒ 2x = 221 ⇒ x = 220
39.
(b) 64

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 Explanation: 64
40. (a) 1625
Explanation:
Indians Foreigners Number of Ways

2 4 6
C2 ×
8
C4 = 1050

3 6 6
C3 ×
8
C6 = 560

4 8 6
C4 ×
8
C8 = 15
Total number of ways = 1625
41.
51!−1
(b) 51!

Explanation: Given, an = n

(n+1)!

50 50

Now, ∑ an = ∑
n

(n+1)!
n=1 n=1

50
(n+1)−1
= ∑
(n+1)!
n=1

50
(n+1) 1
= ∑ { − }
(n+1)! (n+1)!
n=1

50
1 1
= ∑ { − }
n! (n+1)!
n=1

1 1 1 1 1 1 1 1
= ( − ) + ( − ) + ( − ) + … + ( − )
1! 2! 2! 3! 3! 4! 50! 51!

51!−1
=1- 1

51!
=
51!

42.
(c) 12
Explanation: The number of ways in which 5 beads of different colours can be arranged in a circle to form a necklace are (5 -
1)! = 4! = 24
Since the clockwise and anticlockwise arrangement is the same.
⇒ Total number of ways of arranging the beads

= × 4! = × 4 × 3 × 2 = 12
1

2
1

43.
(b) 35
Explanation: 35
44.
(c) 185
Explanation: Required number = 20C2 - 4C2 + 1
= 20×19

2
−
4×3

2
+ 1 = 190 - 6 + 1 = 185

45.
(d) 7
Explanation: 28C2r : 24C2r-4 = 225 : 11
28! 24! 225
⇔ : =
(2r)!(28−2r)! (2r−4)!(24−2r+4)! 11

28! (2r−4)! 225

⇔ × =
(2r)! 24! 11

28×27×26×25 225
⇔ =
2r(2r−1)(2r−2)(2r−3) 11

⇔ 2r(2r - 1) (2r - 2) (2r - 3)

= × 28 × 27 × 26 × 25
11

225

⇔ 2r(2r - 1) (2r - 2) (2r - 3)

= 11 × 28 × 3 × 26
= 11 × 14 × 2 × 3 × 13 × 2
= 14 × 13 × 12 × 11

8 / 13
 By comparing, we get 2r = 14
⇔ r = 7

46.
(b) 12
Explanation: Number of diagonal = 54
n(n−3)
⇒
2
= 54
⇒ n2 - 3n - 108 = 0
⇒ n2 - 12n + 9n - 108 = 0
⇒ n = (n - 12)+ 9 (n - 12) = 0

⇒ n = 12, -9

⇒ n = 12 (∵ n ≠ -9)

47.
(c) 72
Explanation: 72
48.
(c) 2n - 2
Explanation: Let B1 and B2 be two boxes
Each object has two options (B1 or B2) to go.

⇒ There are 2 × 2 × … × 2 = 2n ways to place n objects


n times

These 2n cases include two cases in which either of the boxes remains empty.
⇒ Required number = 2n - 2
49.
(c) 1

2
nan
n
1
Explanation: Given, an = ∑ n
Cr
...(i)
r=0
n

Let bn = ∑ n
r

Cr
r=0

Then, bn = n
0

C0
+ n
1

C1
+ n
2

C2
+ … + n
n

Cn
...(ii)
Also,
bn = n
n
+
n−1

n
+
n−2

n
+ … +
n
0
...(iii) ...[∵ nCr = nCn-r]
C0 C1 C2 Cn

Adding (ii) and (iii), we get

2bn = + n
n
+
C0
+ … + n
n

C1
n
n

C2
n
n

Cn

⇔ 2bn = n( n
1

C0
+ n
1

C1
+ n
1

C2
+ … + n
1

Cn
)

n
r
⇔ 2bn = n ∑ n
Cr
r=0

⇔ 2bn = nan ...[From (i)]

⇔ bn = 1

2
nan

50. (a) 1 + (52)!

Explanation: We know, (r + 1) rPr-1 = (r + 1) r!

1!
= (r + 1)!

So, (2 1P0 - 3 2P1 + ... 51 terms) + (1! - 2! + 3! - ... upto 51 terms)

= [2! - 3! + 4! -... + 52!] + [1! - 2! + 3! - ... + 51!]
= 52! + 1! = 52! + 1
51.
(d) 255
5!
Explanation: Number of words starting with A and having 2T, 1R, 1G, 1E = 2!
= 60
Number of words starting with E and having 2T, 1R, 1A, 1 G = 5!

2!
= 60
5!
Number of words starting with G and having 2T, 1R, 1A, 1E = 2!
= 60

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 Number of words starting with R and having 2T, 1A, 1G, 1E = 5!
= 60
2!

Number of words starting with TAE and having 1T, 1R, 1G = 3!

Number of words starting with TARE and having 1G, 1T = 2!
Total number of words before TARGET = 60 + 60 + 60 + 60 + 6 + 6 + 2 = 254
⇒ Rank of word TARGET is 255.

52.
(d) 5 × 6!
Explanation: 4 boys and 2 girls in circle
⇒ 5! × × 2! ⇒ 5 × 6!
6!

4!2!

53. (a) 25
Explanation: 25
54. (a) 423360
Explanation: If we keep D’s aside, then there are 8 letters of which 2 B’s, 2 C’s and 2 E’s are alike
Number of ways to arrange these 8 letters
8!
= 2!2!2!

= 5040
8 letters create 9 gaps for D’s to be arranged
⇒ The number of ways to do so

= 9C3 = 84 ...[∵ all D's are alike]

⇒ The required number of ways = 5040 × 84
= 423360
55. (a) 8
Explanation: Number of five digit numbers with 2 at 10th place
= 8 × 8 × 7 × 6 = 2688
∵ It is given that, number of five digit number with 2 at 10th place = 336k
∴ 336 k = 2688 ⇒ k = 8

56.
(c) 374
Explanation:
Using the digits 0, 1, 3, 7, 9
number of one digit natural numbers that can be formed = 4
number of two digit natural numbers that can be formed = 20,

(∵ 0 can not come in 1st box)

number of three digit natural numbers that can be formed = 100

and number of four digit natural numbers less than 7000, that can be formed = 250

(∵ only 1 or 3 can come in 1st box)

∴ Total number of natural numbers formed = 4 + 20 + 100 + 250 = 374

57. (a) 21 × 5!
Explanation: There are 3E’s, out of which two are fixed at first and last position. After fixing two E’s 7 letters are left which
can be permuted in 7!

2!
ways i.e., 21 × 5! ways
⇒ The required number of permutations
= 21 × 5!

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58.
(d) 7
Explanation: 7
59.
(b) 1380
Explanation:
Case I: Four-digit numbers greater than 3000

1st place can be filled (by 3 or 4 or 5) in 3 ways. 2nd, 3rd, 4th places can be filled in (by any of remaining 5 digits) 5P3 ways

The number of four-digit numbers = 3 × 5P3

= 180
Case II: Five-digit numbers
The number of five-digit numbers
= 5 × 5 × 4 × 3 × 2 = 600
Case III: Six digit numbers
The numbers of six-digit numbers
= 5 × 5 × 4 × 3 × 2 × 1 = 600
Total number of numbers = 180 + 600 + 600 = 1380

60.
(d) 120
Explanation: Given there are three boxes, each containing 10 balls labelled 1, 2, 3, ... , 10
Now, one ball is randomly drawn from each boxes, and ni denote the label of the ball drawn from the ith. box, (i = 1,2,3).
Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is same as selection of 3 different numbers

from numbers {1, 2, 3, ... , 10} = 10C3 = 120

61. 60
Explanation:
The unit place can be filled by any one of the digits 1, 2, 3, 4 and 5. So the unit place can be filled in 5 ways. Now four digits left.
So the tens place can be filled in 4 ways. The hundreds place can be filled in 3 ways by the remaining 3 digits because repetition
of digits is not allowed.
∴ Total number of 3-digits numbers = 3 × 4 × 5 = 60 .

62. 60
Explanation:
Given digits are 2, 3, 4, 5, 6 and 0.
Any number between 400 and 1000 will be of three digits.
Now, the number should be greater than 400.
So hundred’s place can be filled up by anyone of the digits 4, 5 and 6 in 3 ways.
∴ k = 3

Remaining two places can be filled up by the remaining five digits in 5P2 ways.
Hence, required numbers 3 × 5
P2 = 3 ×
5!

3!
= 60
63. 60480
Explanation:
Required number of ways = 9
P6
9! 9!
= =
(9−6)! 3!

= 9 × 8 × 7 × 6 × 5 × 4 = 60480
64. 35
Explanation:

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 There are 9 courses and number of courses to be selected are 5 in which 2 specific courses are compulsory.
We have to select 3 courses out of remaining 7 courses.
∴ Number of ways of selection = C
7
3

7!

3! 4!
=
7 × 6 × 5 × 4!

3 × 2 × 1 × 4!
= 35
65. 3003
Explanation:
We have,
13C + 13C = 14C [∵ nC + nC n + 1C
6 5 6 r r-1= r
14! n n!
= [∵ Cr = ]
6!(14 − 6)! r!(n − r)!

14! 14 × 13 × 12 × 11 × 10 × 9 × 8!
= =
6! × 8! 6 × 5 × 4× 3× 2× 1× 8!

= 7 × 13 × 11 × 3 = 3003
66. 100
Explanation:
There are total 5-digits. In which, 0 cannot be placed at the hundred's place as in that case, the number will not be of 3-digit. Thus,
hundred’s place can be filled up by any of remaining 4-digits in 4 ways.
Since the digit may be repeated any number of times, so each of the remaining two places can be filled up any of 5-digits in 5
ways each.
Thus, the total number of such arrangements = 52 = 25
Hence, total number of 3-digit numbers = 4 × 25 = 100
67. 1440
Explanation:
There are 8 letters in the word EQUATION including 5 vowels and 3 consonants. Now 5 vowels can be arranged in 5! Ways and 3
consonants can be arranged in 3! Ways. Also the two groups of vowels and consonants can be arranged in 2! Ways.
∴ Total number of permutation = 5! × 3! × 2!

= 120 × 6 × 2 = 1440

68. 120
Explanation:
Here, the total number of children
= 3 boys + 2 girls
=5
The number of ways of filling the first place in group photograph is 5 ways.
Since the first place is filled, then the second place can be filled in 5 -1 = 4 ways.
Similarly, we can find the number of ways of filling third, fourth and fifth place are 3, 2 and 1 ways, respectively.
Hence, by using the fundamental principle of multiplication, all five places can be filled in
5 × 4 × 3 × 2 × 1 = 120 ways

69. 5040
Explanation:
Here the unit place can be filled by any one of the first 10 letters of the English alphabet. So the unit place can be filled in 10
ways. The tens place can be filled in 9 ways by remaining 9 letters of the English alphabet. The hundreds place can be filled in 8
ways by remaining 8 letters of the English alphabet.
The thousands place can be filled in 7 ways by the remaining 7 letters of the English alphabet.
∴ Total number of 4-letter code numbers = 7 × 8 × 9 × 10 = 5040.

70. 778320
Explanation:
There are 4 aces and 48 other cards in a deck of 52 cards. We have to select 1 ace out of 4 aces and 4 other cards out of 48 other
cards.
∴ Number of ways of selection =
4 48
C × C 1 4

48 × 47 × 46 × 45
= 4 ×
4 × 3× 2× 1
= 778320
71. 514.0
Explanation:
Divisible by 2 is 899

2
≈ 450
899
Divisible by 3 is 3
≈ 300

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 Divisible by 7 is 899

7
≈ 128
Divisible by 2 & 7 is 899

14
≈ 64
Divisible by 3 & 7 is 899

21
≈ 43
899
Divisible by 2 & 3 is 6
≈ 150
899
Divisible by 2, 3 & 7 is 42
≈ 21
∴ Total numbers = 450 + 300 - 150 - 64 - 43 + 21 = 514
72. 17.0
Explanation:
No. of ways of selecting 3 boys and 2 girls
= C × C = 168
b
3
g
2

b(b - 1)(b - 2) (g) (g - 1) = 8 × 7 × 6 × 3 × 2

b + 3g = 17
73. 135
Explanation:
Select any 4 correct questions in 6C4 ways.
Number of ways of answering wrong question = 3
∴ Required number of ways = 6C4(1)4 × 32 = 135
74. 31.0
Explanation:
Exponent of 3 in 66! is
66 66 66
⇒ [ ] + [ ] + [ ]
3 9 27

22 + 7 + 2 = 31
75. 5
Explanation:
Given, T n =
n
C3 ⇒ Tn+1 =
n+1
C3

∴ Tn+1 − Tn =
n+1
C3 −
n
C3 = 10 [given]
n n n n n n+1
⇒ C2 + C3 − C3 = 10 . . . [∵ Cr + Cr+1 = Cr+1 ]

nC = 10
⇒ 2
⇒ n=5

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