nC2-nC4=44 find the value of n
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2nC3 = 2n! / (2n - 3)! 3!
nC2 = n! / (n - 2) 2!
2n! / (2n - 3)! 3! X (n - 2)! 2! / n! = 44 / 3
2n (2n-1) (2n-2) (2n-3)! / (2n-3)! 3 x 2! X (n-2)!2! / n(n-1)(n-2)! = 44 /3
2 (2n -1) 2(n-1) / (n-1) = 44
4(2n-1) = 44
2n-1 = 11
2n = 12
n = 6
nC2 = n! / (n - 2) 2!
2n! / (2n - 3)! 3! X (n - 2)! 2! / n! = 44 / 3
2n (2n-1) (2n-2) (2n-3)! / (2n-3)! 3 x 2! X (n-2)!2! / n(n-1)(n-2)! = 44 /3
2 (2n -1) 2(n-1) / (n-1) = 44
4(2n-1) = 44
2n-1 = 11
2n = 12
n = 6
Devashish2000:
thanks brother
but the question is nc2-nc4=44....but never mind...u helped me
but the question is nc2-nc4=44....but never mind...u helped me
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