Statistics: binomial - how to figure out xCx method
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Statistics: binomial - how to figure out xCx method

[From: ] [author: ] [Date: 13-02-03] [Hit: ]
..and then divide by the remaining numbers...7C5 = 21-nCm = m!......
so my textbook is giving me the answers but not showing the steps on how to get the answer....can someone please explain to me showing the step of how they got this?

7C2 = 21 --> i thought initially you multiply down like 7*6*5*4...and then divide by the remaining numbers...for instance 2 down
7C3 = 35
7C4 = 35
7C5 = 21

-
nCm = m!/(n! * (m-n)!)

7C2 = 7!/(2! * 5!)

= 7!/(5! x 2!)
= 7 * 6 * 5 * 4 ...* 1 /(5 * 4 * 3 * 2 * 1 * 2 * 1)
= 7 * 6/2
= 21

All the same, use the same formula for nCm
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