Question

solve this permutation question​

Answer 1

Answer:

7

Step-by-step explanation:

22 P r+1 = 22 × 21 ×20! / (21-r)(20-r)(19-r)(18-r)!

= [(22 × 21)/ (21-r)(20-r)(19-r)] × [20! / (18-r)!]

= = [(22 × 21)/ (21-r)(20-r)(19-r)] × [20 P r+2]

=

now.....22 P r+1 / 20 P r+2 = 11/52

[(22 × 21)/ (21-r)(20-r)(19-r)] = 11/52

[(21-r)(20-r)(19-r)] = 21 × 2 × 52 = 7 × 3 × 2 × 13 × 4 = 14 × 13 × 12 = [(21-7)(20-7)(19-7)]

hence r = 7

Answer 2

Answer- The above question is from the chapter 'Permutations and Combinations'.

Concept used: 1) Formula of permutation is as follows:

$^nP_r = \dfrac{n!}{(n \ - \ r)!}$

2) n! = n.(n - 1).(n - 2)!

Given question: If $\bf ^{22}P_{r+1} : ^{20}P_{r +2} = 11:52, \ find \ r.$

Solution: $^{22}P_{r+1} : ^{20}P_{r+2} = 11:52$

$\dfrac{^{22}P_{r+1}}{^{20}P_{r+2}} = \dfrac{11}{52}\\\\\\\dfrac{22!}{(22 - r - 1)!} \times \dfrac{(20 - r - 2)!}{20!} = \dfrac{11}{52}$

$\dfrac{22 \times 21 \times 20!}{(21 - r)!} \times \dfrac{(18 - r)!}{20!} = \dfrac{11}{52}$

$\dfrac{2 \times 21 \times (18-r)!}{(21-r)(20-r)(19-r)(18-r)!} = \dfrac{1}{52}$

(21 - r)(20 - r)(19 - r) = 2 × 21 × 52

(21 - r)(20 - r)(19 - r) = 2 × 3 × 7 × 2 × 2 × 13

(21 - r)(20 - r)(19 - r) = 14 × 13 × 12

⇒ 21 - r = 14

r = 21 - 14

r = 7

$\tt \boxed{\therefore \ r = 7}$