# Permutations & Combinations ?4C2 ? 11C5?

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### 1 Answer

You should use factorial formula for combinations such that:

`C_n^k = (n!)/(k!(n-k)!)`

Reasoning by analogy yields:

`C_4^2 = (4!)/(2!(4-2)!) => C_4^2 = (4!)/(2!2!)`

You may write 4! such that:

`4! = 1*2*3*4 = 2!*3*4`

`C_4^2 = (2!*3*4)/(2!2!) => C_4^2 = (3*4)/(2!) => C_4^2 = 12/2`

`C_4^2 = 6`

You may evaluate `C_11^5 ` using the above factorial formula such that:

`C_11^5 = (11!)/(5!(11-5)!) => C_11^5 = (11!)/(5!*6!)`

`C_11^5 = (6!*7*8*9*10*11)/(1*2*3*4*5*6!)`

`C_11^5 = (7*3*2*11) => C_11^5 = 462`

**Hence, evaluating the given combinations using factorial formula yields `C_4^2 = 6` and `C_11^5 = 462` .**