Q.If $^nC_4={}^nC_6$, find $^{12}C_n$.

A handshake is an unordered pair, so the answer is $\binom{13}{2}=78$.

A handshake happens between two guests, and the pair $\{A,B\}$ is the same as $\{B,A\}$ — order does not matter — so we count combinations, not permutations.

The number of handshakes among 13 guests is …

$^{100}C_{99}=100$.

Use the property $^nC_r={}^nC_{n-r}$, which lets us replace a large lower index by a small one:

$^{100}C_{99}={}^{100}C_{100-99}={}^{100}C_{1}.$ …

Choosing 4 players from 5 is $\binom{5}{4} = 5$, so the answer is option (c).

Selecting a team is a combination problem — the order of selection does not matter. The number of ways to choose $r$ objects from $n$ is

$\binom{n}{r} = \frac{n!}{r!\,(n-r)!}.$

Here $n = 5$ and $r = 4$:

…