Re: \"choose\"- another question from a math illiterate
Just to add one more comment: the 'C' actually stands for 'combination'. It's like you wrote - you're choosing a number of cards. The key is that order doesn't matter. (Since AK is the same as KA, for example.) If we were picking 5 cards in order, there would be 52*51*50*49*48 ways. (One less card as you go.) Since order doesn't matter, we have to divide by the number of ways of ordering those 5 cards: 5*4*3*2*1. (5 choices for the first card, 4 for the 2nd, etc.)
Hence C(52,5) = (52*51*50*49*48)/(5*4*3*2*1) as others have already mentioned.
Nothing new to add for the calculation - just wanted to maybe give some context for it.
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