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If .^(n)C(8) = .^(n)C(2), find .^(n)C(2)...

If `.^(n)C_(8) = .^(n)C_(2)`, find `.^(n)C_(2)`.

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To solve the problem where \( \binom{n}{8} = \binom{n}{2} \), we can follow these steps: ### Step 1: Use the property of combinations We know that \( \binom{n}{r} = \binom{n}{n-r} \). In this case, we have: \[ \binom{n}{8} = \binom{n}{2} \] This implies that: ...
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STATEMENT - 1 : If .^(2n+1)C_(1) + .^(2n+1)C_(2)+………+ .^(2n+1)C_(n)= 4095 , then n = 7 STATEMENT - 2 : .^(n)C_(r )= .^(n)C_(n-r) where n epsilon N, r epsilon W and n ge r

The A.M. of the series .^(n)C_(0), .^(n)C_(1), .^(n)C_(2),….,.^(n)C_(n) is