Question

In a hockey series between teams $X$ and $Y$, they decide to play till a team wins a $m$ match. Then the number of ways in which team $X$ wins:
A. ${{2}^{m}}$
B. $^{2m}{{P}_{m}}$
C. $^{2m}{{C}_{m}}$
D. None of these

Answer 1

Hint: We see that X can win the series by losing 0 match in any of the first $m-1$ matches and winning the ${{m}^{th}}$ match in ${}^{m-1}{{C}_{0}}$ way or losing 1 match in any of the first $m$ matches and winning the ${{\left( m+1 \right)}^{th}}$ match in ${}^{m}{{C}_{1}}$ way or so on. We find the maximum number of matches X and Y have to play as $2n-1$ so that X can win the series with $m$ wins. We use rule of sum and find the total number of ways so that X can win the series as ${}^{m-1}{{C}_{0}}+{}^{m}{{C}_{1}}+{}^{m+1}{{C}_{2}}+...+{}^{2m-2}{{C}_{m-1}}$. We use the hockey-stick identity $\sum\limits_{i=r}^{n}{^{i}{{C}_{r}}}={}^{n+1}{{C}_{r+1}}$ to find the answer. \[\]

Complete step by step answer:
We know from Hockey-stick identity of combinatorial mathematics that
\[\sum\limits_{i=r}^{n}{^{i}{{C}_{r}}}={}^{n+1}{{C}_{r+1}},\text{ for }n,r\in \mathsf{\mathbb{R}},\mathsf{}n\ge r\]
We are given the question that a hockey series between team $X$ and $Y$, they decide to play till a team wins $m$ matches. We see here that the last match in the series have to be won by X if we want X to win the series and the series will end after that match and the win of that match will be ${{m}^{\text{th}}}$ win for X since two teams will be keep playing until one of them wins $m$ matches.
So we here see that there can be maximum $m+\left( m-1 \right)=2m-1$matches between them where X wins $m$ matches and loses $m-1$matches alternatively as
\[WLWL.....\underline{W}\leftarrow {{m}^{\text{th}}}\text{ win for }X\]
We also see that in the minimum case X and Y have to play $m$ matches and X does not lose any match.
\[WWW...\left( m\text{ times} \right)...W\leftarrow {{m}^{\text{th}}}\text{ win for X}\]
In the minimum case X has to have 0 losses in first $m-1$ matches in ${}^{m-1}{{C}_{0}}$ ways and then win the ${{m}^{\text{th}}}$ match to win the series. \[\]
If X and Y play $m+1$ matches then X can win series by losing any 1 of the match in first $m$ matches in ${}^{m}{{C}_{1}}$ ways and then win the ${{\left( m+1 \right)}^{\text{th}}}$ match.\[\]
If X and Y play $m+2$ matches then X by losing any 2 matches in first $m+1$ matches ${}^{m+1}{{C}_{2}}$ ways and then win the ${{\left( m+2 \right)}^{\text{th}}}$ to win the series.\[\]
We can continue like this by increasing the number of matches until we reach the maximum $\left( 2m-1 \right)$. If X and Y play $2m-1$ matches then X by losing any $m-1$ matches in first $2m-2$ matches ${}^{2m-2}{{C}_{m-1}}$ ways and then win the ${{\left( 2m-1 \right)}^{\text{th}}}$ match to win the series.\[\]
 So by rule of sum X can the number of ways X can win the series with $m$ wins is;
\[\Rightarrow {}^{m-1}{{C}_{0}}+{}^{m}{{C}_{1}}+{}^{m+1}{{C}_{2}}+...+{}^{2m-2}{{C}_{m-1}}\]
We use us the combinatorial formula ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ in the above step to have;
\[\Rightarrow {}^{m-1}{{C}_{m-1}}+{}^{m}{{C}_{m-1}}+{}^{m+1}{{C}_{m-1}}+...+{}^{2m-2}{{C}_{m-1}}\]
We use the hockey stick identity for $r=m-1$ and $n=2m-2$ and have
\[\Rightarrow {}^{2m-2+1}{{C}_{m-1+1}}={}^{2m-1}{{C}_{m}}\]
So the number of ways X wins the series with $m$ wins is ${}^{2m-1}{{C}_{m}}$

So, the correct answer is “Option D”.

Note: We note to be careful that we are taking combinations on the number of losses not on number of wins. We also note that the question assumes no match ends in a draw. The hockey-stick identity comes from a diagonal of Pascal’s triangle. The rule of sum states that if there are $m$ ways to do one thing and $n$ ways to do another thing then we can do either of things in $m+n$ ways.