Permutation and Combination - Algebra -Integer Type Questions(JEE ADVANCED) Part 1

 Q.1

Correct Answer is: [7704]  

Solution Explain:

Q.2

Correct Answer is: [4373]  

Solution Explain:

Q.3

Correct Answer is: [15]  

Solution Explain:

Q.4

Correct Answer is: [18]  

Solution Explain:

Q.5

Correct Answer is: [160]  

Solution Explain:

Q.6

Correct Answer is: [625]  

Solution Explain:

Q.7

Correct Answer is: [5]  

Solution Explain:

First 6 distinct digits can be selected in 10C6 ways. Now the position of smallest digit in  them is fixed i.e position 4. Of the remaning 5 digits, 2 can be selected  in 5C2 ways. These two digits can be placed to the right of 4th position in one  way only. The remaning three digits to the left  of 4th position are  in the required order automatically. So n(S) = 10C6 × 5C2 = 210 × 10 =2100

Q.8

Correct Answer is: [128]  

Solution Explain:

Q.9

Correct Answer is: [924]  

Solution Explain:

Q.10

Correct Answer is: [32]  

Solution Explain:

Q.11

Correct Answer is: [720]  

Solution Explain:

Q.12

Correct Answer is: [8]  

Solution Explain:

7!= 24×32 ×5×7
Since the factor should be odd as well as of  the form 3t + 1, the factor cannot be a multiple of either 2 or 3. So the factors may be1,5,7 and 35 of which  only 1 and 7 are of the from 3t + 1 whose sum is 8

Q.13

Correct Answer is: [126]  

Solution Explain:

Q.14

Correct Answer is: [6]  

Solution Explain:

Here x1 x2 x3 = 22 × 3 ×5. Let number of twos given  to each of  x1 x2 x3 be a, b,c. Then a+b+c=2, a,b,c ≥ 0
The number of integral  solutions of this equations is equal  to coefficent of X2 in (1-x)-3 i e 4C2 i.e the available  2 twos can be distributed among x1 x2 x3 in 3C2 = 3 ways (=coefficent of X in (1-x)-3)
∴ Total  number of ways = 4C2 × 3C2 × 3C2 = 6 ×3×3 =54 ways

Q.15

Correct Answer is: [7]  

Solution Explain:

Q.16

Correct Answer is: [1560]  

Solution Explain:

Q.17

Correct Answer is: [309]  

Solution Explain:

Q.18

Correct Answer is: [77]  

Solution Explain:

Q.19

Correct Answer is: [36]  

Solution Explain:

Q.20

Correct Answer is: [4]  

Solution Explain:

If three numbers are in G.P., then their exponent must be in A.P.

If a, b, c are selected number in G.P., then the exponents of a and c both are either odd or both even, or otherwise exponent b will not be integer.

Now two odd exponent (from 1, 2, 3, …, 10) can be selected in 5C2 ways and two even exponent can be selected in 5C2 ways.

Hence number of G.P.’s are 25C2 = 20

Q.21

Correct Answer is: [6]  

Solution Explain:

Q.22

Correct Answer is: [490]  

Solution Explain:

Q.23

Correct Answer is: [16]  

Solution Explain:

Q.24

Correct Answer is: [80]  

Solution Explain:

Q.25

Correct Answer is: [3]  

Solution Explain:

Q.26

Correct Answer is: [3720]  

Solution Explain:

Q.27

Correct Answer is: [625]  

Solution Explain:

Q.28

Correct Answer is: [6]  

Solution Explain:

Q.29

Correct Answer is: [777]  

Solution Explain:

Q.30

Correct Answer is: [30]  

Solution Explain:

Q.31

Correct Answer is: [288]  

Solution Explain:

Let two particular boys as one boy, we have only four boys.
∴5 boys can be seated at a round table when two particular boys are always together = 3! 2!
⇒ 4 girls have are 4 places
∴ 4 girls can be arranged in 4! ways
∴ Required number = 3! 2! 4! = 6. 2. 24
                                             = 288

Q.32

Correct Answer is: [132]  

Solution Explain:

Sum of the digits in the tens place
= (4 – 1)! (4 + 5 + 6 + 7)
= 6 × 22
= 132

Q.33

Correct Answer is: [720]  

Solution Explain:

Q.34

Correct Answer is: [1200]  

Solution Explain:

Q.35

Correct Answer is: [117]  

Solution Explain:

Q.36

Correct Answer is: [6]  

Solution Explain:

Q.37

Correct Answer is: [2]  

Solution Explain:

21,22,23,.........,k-1,k
A.M =21+k/2, G.M = √(21.k)
⇒ k=21 λ2, λ∈I also 100 ≤ k ≤ 999 and k should be odd
⇒ 100/21 ≤ λ2 ≤ 999/21 ⇒ 4.76 ≤  λ2 ≤ 47.57⇒ λ =3,4,5,6 but λ should be  odd ⇒ odd λ =3,5 ⇒ 'k' can assume2 different values

Q.38

Correct Answer is: [288]  

Solution Explain:

Q.39

Correct Answer is: [1251]  

Solution Explain:

Q.40

Correct Answer is: [6]  

Solution Explain:

3630 = 2 × 3× 5 ×112
Now a divisor will be of the form (4n+1) if divisor is form the help of (4n+1) type  number or  by (4n+3) types number taken even times.
Hence divisors are 1,5,3 ×11, 112,5 ×3 × 11. i e, 6