Question 1 :
If (n+1)C8 : (n−3) P4 = 57 : 16, find the value of n.
Solution :
(n+1)C8 = (n + 1)!/(n + 1 - 8)! 8! ==> (n + 1)!/(n - 7)! 8! --(1)
(n−3) P4 = (n - 3)!/(n - 3 - 4)! ==> (n - 3)!/(n - 7)! --(2)
(1) : (2)
[(n + 1)!/(n - 7)! 8!] : [(n - 3)!/(n - 7)! ] = 57 : 16
(n + 1)n(n - 1)(n - 2)/(8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2) = 57 / 16
(n + 1) n (n - 1)(n - 2) = (57/16) ⋅ (8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2)
(n + 1) n (n - 1)(n - 2) = (57 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3)
(n + 1) n (n - 1)(n - 2) = (3 ⋅ 19 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3)
(n + 1) n (n - 1)(n - 2) = 21 ⋅ 20 ⋅ 19 ⋅ 18
n = 20
Hence the value of n is 20.
Question 2 :
Prove that 2nCn = [2n × 1 × 3 × ·· · (2n − 1)] / n!.
Solution :
L.H.S
= 2nCn = 2n!/(2n-n)! n! = 2n!/n! n!
= [2n (2n - 1)(2n - 2) ............. ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1] / n! n!
= [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1) ] [2 ⋅ 4 ⋅ 6 ⋅.......2n] / n! n!
= [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1) ] 2n [1 ⋅ 2 ⋅ 3 ⋅.......n] / n! n!
= [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1) ] 2n n! / n! n!
= 2n [ 1 ⋅ 3 ⋅ 5 ⋅.............(2n - 1)] / n! ---> R.H.S
Hence it is proved.
Question 3 :
Prove that if 1 ≤ r ≤ n then n × (n−1) Cr−1 = (n − r + 1) nCr−1.
Solution :
Question 4 :
Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?
Solution :
Number of ways of selecting 7 players out of 14 players
= 14C7
= 14! / (14 - 7)! 7!
= 14! / 7! 7!
= (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7!)/7! 7!
= (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8) / 7!
= (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8) / (7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2)
= 3432
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