Integer Answer Type Questions for JEE: Permutations & Combinations | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The number of numbers from 1 to 10⁶ (both inclusive) in which two consecutive digits are same is equal to 402128 + K where K is a single digit number then K must be equal to ______.

Ans. 2
 No. of  n digit numbers in which no two consecutive digits are same = 9ⁿ
⇒ no. of numbers from 1 to 10⁶ in which no two consecutive digits are same =
Required number = 10⁶ - 597870 = 402130 = 402128 + 2
K = 2

Q.2. The number of 3 digit odd numbers divisible by 3, which can be formed using the digits 3, 4, 5, 6 when repetition of digits within the number is allowed is 8 + λ then value of λ is 

Ans. 3
Three digits odd numbers using only 3 and only 5 are 2.
Three digit odd numbers using 3, 4 and 5, are 4.
Three digit odd numbers using 2, 3 and 1, 6 are 2.
Three digit odd numbers using two 6 and one 3 are 1.
Three digit odd numbers using two 3 and one 6 are 2.
So, total three digit numbers = 11.

Q.3. Let n be the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9, no digit being repeated. Find the value of n/32.

Ans. 6
A five digit integer is always greater than 7000.
The number of such integers is 5P5 = 5! = 120.
For a four digit integer to be greater than 7000, it must begin with 7, 8, or 9.  
The number of such integers is 120 + 72 = 192.

Q.4. If number of numbers greater than 3000, which can be formed by using the digits 0, 1, 2, 3, 4, 5 without repetition, is n then n / 230 is equal to

Ans. 6
No. of 4 digit numbers = 3 × 5 × 4 × 3 = 180
No. of 5 digit numbers = 5 × 5 × 4 × 3 × 2 = 600
No. of 6 digit numbers = 5 × 5 × 4 × 3 × 2 = 600
n = 1380
⇒ n/230 = 6

Q.5. If one test (on screening paper basis) was conducted on Batch A, maximum number of marks is (90 × 3) = 270. 4 students get the marks lower than 80. Coaching institute decide to inform their guardians, that is why their result card were sent to their home. The number of ways, in which all the letters were put in wrong envelopes, is

Ans. 9
The number of ways in which all the letters are in wrong envelopes
=

Q.6. Let a be a factor of 120. Then the number of positive integral solutions of x₁ , x₂ , x₃ = a is k then k / 80 is

Ans. 4
Let x₄ be such that
Then the number of positive integral solutions of x₁x₂ x₃ = a is the same as that of the number of positive integral solutions of x₁ x₂ x₃ x₄ = 120 = 2³ x 3 x 5 We can assign 3 and 5 to unknown quantities in 4 x 4 ways. We can assign all 2 to one unknown in ⁴C₁ ways, to two unknown in ( ⁴C₂ ) (2) and to three unknown in ⁴C₃ ways. Hence the number of required solutions
=
= 4 x 4 x 20 = 320 = 0320
Then, k = 4

Q.7. Find the number of different necklaces that can be made from 17 identical pearls and 2 different diamonds.

Ans. 9
The nineteen pearls (17 identical + 2 different) can be arranged in a circle in 18!/17! =18 ways Now, we can see that mutual arrangement of two different diamonds does not give a different necklace (merely gives a minor image). Hence, the required number of different necklaces is 18/2 = 9

Q.8. The exponent of 7 in ¹⁰⁰C₅₀ is:

Ans. 0

Exponent of 7 in 100 ! = 16
Exponent of 7 in 50!
Exponent of 7 in (50!)2 = 16
∴ Exponent of 7 in ¹⁰⁰C₅₀ = 16 - 16= 0 .

Q.9. Nine hundred distinct n-digit number s are to be formed using exactly the three digits. 2, 5 and 7. The smallest value of n for which this is possible is.

Ans. 7
For n = 6
3 x 3 x 3 x 3 x 3 x 3=729<900
For n = 7
3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187 > 900
For n = 8
Number of n-digits formed > 900
Since the least n is required.
∴ n = 7.

Q.10. The number of ways of arranging 11 objects A, B, C, D, E, F ,α ,α ,α , β , β so that every b lie between two a (not necessarily adjacent) is K x  6! x ¹¹C₅ , then K is ____.

Ans. 3
There are three major ways ααbbα ,αββαα and αβαβα Each major way has six empty spaces. The number of ways of putting letters at these empty spaces must be nonnegative integer function of x₁ + x₂ + ... + x₆ = 6
=
No. of arrangements is = 3 x ¹¹C₅  x  6! ⇒ K + 3