Why it is $k-1$ elements of $S$, instead of $T$ here? Since when $a$ is removed, there is $k-1$ elements left in $T$?
Why it is $k-1$ elements of $S$, instead of $T$ here? Since when $a$ is removed, there is $k-1$ elements left in $T$?
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discrete-mathematics
1 Answers
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The $k-1$ other elements of your set with $k$ elements must not be equal to $a$, since your set consists of $a$ together with these $k-1$ other elements. That is, they must all be elements of the set $S=T-\{a\}$. If you allowed them to be elements of $T$ rather than $S$, that would mean it is possible for them to be equal to $a$.