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Question: In a hockey series between teams \(X\) and \(Y\), they decide to play till a team wins a \(m\) match...

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

Explanation

Solution

We see that X can win the series by losing 0 match in any of the first m−1m-1 matches and winning the mth{{m}^{th}} match in m−1C0{}^{m-1}{{C}_{0}} way or losing 1 match in any of the first mm matches and winning the (m+1)th{{\left( m+1 \right)}^{th}} match in mC1{}^{m}{{C}_{1}} way or so on. We find the maximum number of matches X and Y have to play as 2n−12n-1 so that X can win the series with mm wins. We use rule of sum and find the total number of ways so that X can win the series as m−1C0+mC1+m+1C2+...+2m−2Cm−1{}^{m-1}{{C}_{0}}+{}^{m}{{C}_{1}}+{}^{m+1}{{C}_{2}}+...+{}^{2m-2}{{C}_{m-1}}. We use the hockey-stick identity ∑i=rniCr=n+1Cr+1\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
∑i=rniCr=n+1Cr+1, for n,r∈R,n≥r\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 XX and YY, they decide to play till a team wins mm 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 mth{{m}^{\text{th}}} win for X since two teams will be keep playing until one of them wins mm matches.
So we here see that there can be maximum m+(m−1)=2m−1m+\left( m-1 \right)=2m-1matches between them where X wins mm matches and loses m−1m-1matches alternatively as
WLWL.....W‾←mth win for XWLWL.....\underline{W}\leftarrow {{m}^{\text{th}}}\text{ win for }X
We also see that in the minimum case X and Y have to play mm matches and X does not lose any match.
WWW...(m times)...W←mth win for XWWW...\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−1m-1 matches in m−1C0{}^{m-1}{{C}_{0}} ways and then win the mth{{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+2m+2 matches then X by losing any 2 matches in first m+1m+1 matches m+1C2{}^{m+1}{{C}_{2}} ways and then win the (m+2)th{{\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 mm wins is;
⇒m−1C0+mC1+m+1C2+...+2m−2Cm−1\Rightarrow {}^{m-1}{{C}_{0}}+{}^{m}{{C}_{1}}+{}^{m+1}{{C}_{2}}+...+{}^{2m-2}{{C}_{m-1}}
We use us the combinatorial formula nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}} in the above step to have;
⇒m−1Cm−1+mCm−1+m+1Cm−1+...+2m−2Cm−1\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−1r=m-1 and n=2m−2n=2m-2 and have
⇒2m−2+1Cm−1+1=2m−1Cm\Rightarrow {}^{2m-2+1}{{C}_{m-1+1}}={}^{2m-1}{{C}_{m}}
So the number of ways X wins the series with mm wins is 2m−1Cm{}^{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 mm ways to do one thing and nn ways to do another thing then we can do either of things in m+nm+n ways.