Number of ways in which two elephants can be chosen out of $n$ elephants. Question:
In how many ways can two elephants be chosen out of $n$ elephants ?
Answer:
Clearly, it is  $^nC_2$.
My solution:
Let's choose an elephant . This can be done in $^nC_1$ ways. Now, to complete the selection , we have to take one more elephant out of left $(n-1)$ elephants. This can be done in $^{n-1}C_1$ ways. Hence, by multiplication rule of counting, total number of ways = $(^nC_1)  . (^{n-1}C_1)$ .
Problem:
However, $^nC_2≠ (^nC_1)  . (^{n-1}C_1)$ . So , where did I go wrong ?
Thanks. I am a beginner in permutations and combinations so forgive if I am doing a very basic mistake.
 A: Hello this is actually a good question if you are a beginner, the way I solved your question is, just assume 4 elephants (n=4).
$\\$No. of ways you can select 2 elephants = $^{4}C_{2} = 6$ ways.
Now let's solve this question by the way you told, acc. to you, first we select one elephant, that is $^{4}C_{1}$, now select another elephant out of remaining (3) elephants, so that'll be $^{3}C_{1}$, so total no. of ways we can select elephant = $^{4}C_{1}\cdot^{3}C_{1} = 12$ ways.
Let's analyze the mistake we did here, let's name the elephants E1, E2, E3 and E5.
Case-1: Let's select E2 first and than E3.
Case-2: Let's select E3 first and than E2.
See, in both the cases, we've actually selected 2 elephants but order differed there, but technically we don't see the order in selection process, and so both are same ways. and that's why 2 cases have been clubbed into 1, that's what the mistake we did in selection in two steps (the method that you told), what would have happened while selection is in one case we might have selected E2 first and than E3, and in another case we might have selected E3 first and than E2, and both are same things because we're just doing selections. so yeah this thing would have happened with all the cases formed, so we've to divide our answer by 2 to get those all same cases clubbed into one. Hope you might have got what I wanted to say.
Let's solve the mathematical equation now that you told.
$^{n}C_{2} = \frac{({n}C_{1}\cdot^{n-1}C_{1})}{2}
\\ \frac{n!}{2! \cdot(n-2)!} = \frac{(n)(n-1)}{2}
\\ \frac{(n)(n-1)}{2}=\frac{(n)(n-1)}{2}
\\ LHS = RHS$
I hope you might have got why'd we divided by 2 in your method.
