1.

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

Answer» `.^(n)C_(x) = .^(n)C_(Y)`
`rArr n = x +y`
`:. .^(n)C_(8) = .^(n)C_(2)`
`rArr n = 8 +2 = 10`
Therefore, `.^(n)C_(2) = .^(10)C_(2) = (10!)/(2!8!) = (10 xx 9 xx 8!)/(2xx 1 xx 8!) = 45`.


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