If n C 4- n C 5 and n C 6 are in A-P- then the value of n can beA- 14B- 11C- 7D- 8

The correct options are A 14 C 72.^n C_5 = ^n C_4 + ^n C_6 2. n!/(n 5)!5! = n!/4!(n 4)! + n!/(n 6)!(6!) On solving (17 n)(n 4) = 30 n = 7, 14 ...

If n C 4, n C...

Question

If $^{n} C_{4},^{n} C_{5} a n d^{n} C_{6}$ are in A.P., then the value of n can be

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^{n} C_{4},^{n} C_{5}

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