ヤ€ 2011 1 ャ€ソ€Вｇ€
㈤1¤ㄦ1ヨ1€1ヨㄦ €В m
$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 lt;k lt; n$. A reason that we do define
.1 やュ﹂€ ャ€
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈢-1. ぇ ...
.PUц╂€CPUцㄦВ〃€
.8.1 2025DIYㄨ ㄨユ€ту
.ф€おс€ ㄨ уtx 5090d
€╂ｇ€€ぞ€ㄥф
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈤1¤ㄦ1ヨ1€1ヨㄦ €В m
$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 lt;k lt; n$. A reason that we do define
.1 やュ﹂€ ャ€
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈢-1. ぇ ...
.PUц╂€CPUцㄦВ〃€
.8.1 2025DIYㄨ ㄨユ€ту
.ф€おс€ ㄨ уtx 5090d
€╂ｇ€€ぞ€ㄥф
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈤1¤ㄦ1ヨ1€1ヨㄦ €В m
$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 lt;k lt; n$. A reason that we do define
.1 やュ﹂€ ャ€
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈢-1. ぇ ...
.PUц╂€CPUцㄦВ〃€
.8.1 2025DIYㄨ ㄨユ€ту
.ф€おс€ ㄨ уtx 5090d
€╂ｇ€€ぞ€ㄥф
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈤1¤ㄦ1ヨ1€1ヨㄦ €В m
$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 lt;k lt; n$. A reason that we do define
.1 やュ﹂€ ャ€
ヤ€ 2011 1 ャ€ソ€Вｇ€
㈢-1. ぇ ...
.PUц╂€CPUцㄦВ〃€
.8.1 2025DIYㄨ ㄨユ€ту
.ф€おс€ ㄨ уtx 5090d
€╂ｇ€€ぞ€ㄥф