tanmay056 wrote:
I am not sure if - 5C2 is going to be correct one .... It seems to me that 5C3 is going to be more precise answer. Because we need to select group of 3 people out of 5. Though both of them will result in 10 . Please clarify of choosing 5C2 over 5C3.
\(5C2 = 5C3\)
Note that \(C^k_n =\frac{n!}{k!(n-k)!}\)
\(C^{n-k}_n=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{(n-k)!k!}\)
Hence \(C^k_n = C^{n-k}_n\)
That's why \(C^3_5 = C^2_5\)
Now let's compare \(5C2\) vs \(5C3\).
If you choose 3 out of 5, there are \(5C3\) different ways.
If you choose 3 out of 5, that means you drop 2 of 5 and 3 left. Thus there are \(5C2\) different ways.
Two different ways with different concept, but have the same result. So you could choose any one which suits you.
I've seen in other questions in which I would have to multiply Combinations or subtract an integer from the Combination. Can you tell me why here we only do 1 Combination?
I think when you need to multiply is when there are, for example, women and men. And to subtract I'm not sure of an instance at the moment.