Practical example for the identity $r(r-1) \binom{m}{r} =m(m-1) \binom{m-2}{r-2}$

Question

I see that many a times we use the result

$$r(r-1) \binom{m}{r} =m(m-1) \binom{m-2}{r-2}.$$

I know it is easy to prove the result, but how to relate it to a practical problem, i.e. how do I look at it as a physical example?

I would highly appreciate if any one can help me relate this result to a practical situation.

I know it may be easy but I am just starting up.

Answer 1

Assuming you are looking for a combinatorial proof of the identity $m(m-1)\binom{m-2}{r-2}=r(r-1)\binom{m}{r}$

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Given $m$ people, we wish to count how many ways we can form a committee with a chairman, a vice-chairman, and $r-2$ lower ranking committee members (for a total of $r$ committee members if you include the chair and vice-chair). We do this by counting in two different methods.

Method 1:

• Pick the chairman from the overall pool of people. $m$ choices
• Pick the vice-chairman from the remaining pool of people. $m-1$ choices
• Pick $r-2$ additional people to serve on the committee. $\binom{m-2}{r-2}$ choices

This gives a total of $m(m-1)\binom{m-2}{r-2}$ ways in which we can accomplish this.

Method 2:

• Pick the $r$ people for the committee. $\binom{m}{r}$ choices
• From those people selected for the committee, pick who will be the chairperson: $r$ choices
• From those people selected for the committee remaining, pick who will be the vice-chairperson: $r-1$ choices

This gives a total of $r(r-1)\binom{m}{r}$ ways in which we can accomplish this.

As both methods give a count for the same scenario, by fundamental principles of counting, the two expressions must be equal.