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Q) Find the number of ways in which India can win the series of 11 matches (If no match is drawn and all matches played).

My answer:${11\choose 6}\cdot2^5$

Answer provided in book:$2^{10}$

My approach For winning a series India must win at least 6 matches of the 11 played which can be done in $11\choose6$ ways and the other 5 matches can be either won or lost which can be done in $2^5$ ways.

Therefore total no. of ways India can win the match is ${11\choose 6}\cdot2^5$.

'''Where am I getting wrong ?'''

DeBARtha
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  • Welcome to MSE. In order for MathJax commands to be effective, they must be enclosed in$ signs. For example, $x^2$ shows up as $x^2$. – saulspatz May 27 '21 at 04:58
  • Note that the number of ways India can win is the same as the number of ways the other team (Australia?) can win, so the answer $2^{10}$ is immediate. To see what you're doing wrong, look at a smaller example, like five matches, where you can write everything out and see where you're double-counting. – Gerry Myerson May 27 '21 at 05:03
  • Just as you mentioned, India needs to win 'at least' 6 matches of the series. So where are the other possibilities, like if India wins 7 or more games? – DeBARtha May 27 '21 at 05:11

1 Answers1

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The mistake in your calculation is that you are double counting. Suppose India wins $7$ matches. You have counted that outcome $7$ times, because there are $7$ ways to choose $6$ of the winning games. The proper calculations is to note that India can win exactly $6$ matches or exactly $7$, etc. so that the total is $$\binom{11}6+\binom{11}7+\binom{11}8+\binom{11}9+\binom{11}{10}+\binom{11}{11}\tag1$$

Since $\binom{11}{k}=\binom{11}{11-k},\ k=0,1,\dots,11$, we see that $(1)$ is equal to $$\frac12\sum_{k=0}^{11}\binom{11}k=\frac{2^{11}}2=2^{10}$$

true blue anil
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saulspatz
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