# Important Questions for Class 11 Chapter 7 - Permutation and Combination

The free set of important questions carries step-by-step explanation and solution. Our math subject experts with years of experience have studied the CBSE exam patterns very carefully and come up with these set of questions based on the current CBSE syllabus. The questions are based on each topic and chapter of the CBSE class 11th syllabus. Download your free PDF now and start practicing!

**Important questions based on NCERT syllabus for Class 11 Chapter 7 - Permutation and Combination:**

*Question 1*: Find r. 5*4p_r = 6*5p_r-1

*Solution*:

Given 5*4p_r = 6*5p_r-1

=> 5 4!/(4-r)! = 6 * 5!/(5-r+1)!

=> 5 4!/(4-r)! = 6 * 5 * 4!/(6-r)!

=> 1/(4-r)! = 6/(6-r)(5-r)(4-r)!

=> (6-r)(5-r) = 6

=> 30 - 6r - 5r +r^2 = 6

=> r^2 -11r + 30 = 6

=> r^2 -11r + 24 = 0

=> r^2 - 8r - 3r + 24 = 0

=> r(r - 8) - 3(r - 8) = 0

=> (r - 8)(r - 3) = 0

=> r = 8 or r = 3

We will not consider r = 8 because if we substitute in the factorial it will be negative.

*Question 2*: There are 40 teachers in the college. in that one principal, one vice principal and one teacher in charge has to be appointed. in how many ways this can be done?

*Solution*:

In 40 way principal, 39 ways vice principla and 38 way one teacher in charge can be appointed.

Hence the number of ways the selection can be made is 40*39*38 = 59,280 ways.

*Question 3*: If P_m stands for mP_m, then prove that 1 + 1P_1 + 2P_2 + 3P_3+.... + nP_n = (n+1)!

Solution: Given If P_m stands for mP_m

If P_m = mP_m then mP_m = m!......(1)

Have to prive that 1 + 1P_1 + 2P_2 + 3P_3+.... + nP_n = (n+1)!

Consider 1 + 1P_1 + 2P_2 + 3P_3+.... + nP_n = 1 + 1P_1 + 2P_2 + 3P_3+.... + nn! [ nP_n = n!(from (1)]

= 1 + summation r = 1 to n (r + 1 - 1) * r!

= 1 + summation r = 1 to n (r + 1)r! - r!

= 1 + summation r = 1 to n (r + 1)! - r!

= 1+ [2! - 1!] + [3! - 2!] + ....... + [(n+1)! - n!]

= 1+(n+1)! - 1

= (n+1)!