Question

In a hockey series between team X and Y, they decide to play till a team wins ‘10’ match. Then the number of ways in which team X wins is $\dfrac{{}^{20}{{C}_{m}}}{2},$ then m equals.

Answer 1

Hint: We solve this question by using the basic combinations formula. We need team X to win and let us assume there are no matches played in the series. According to the given condition, they need to win 9 matches in n-1 matches and then win the nth match. We use this concept to calculate the number of ways in which team X wins.

Complete step by step solution:
In order to solve this question, let us note down what is given to us. It is said that the teams X and Y play a series of matches till one of the teams has 10 wins. We need to count the number of ways for team X to win the series. Let us assume they play n matches. So now, for X to win, they need to win 9 matches out of the n-1 matches and win the nth match.
It is important to note that the value of n cannot be greater than 19 because if it is, that means the team Y wins. This is so because if team X wins 9 matches out of, say 20 matches, this means that team Y has won 11 matches, so they win the series.
Using all the combinations of team X winning in the matches for n ranging from a minimum of 9 matches to maximum of 18 matches, in which they require the 9 wins is given by,
$\Rightarrow {}^{9}{{C}_{9}}+{}^{10}{{C}_{9}}+{}^{11}{{C}_{9}}+{}^{12}{{C}_{9}}+{}^{13}{{C}_{9}}+{}^{14}{{C}_{9}}+{}^{15}{{C}_{9}}+{}^{16}{{C}_{9}}+{}^{17}{{C}_{9}}+{}^{18}{{C}_{9}}$
This is so because to win 9 matches, a minimum of 9 matches needs to be played and if more than 19 matches are played, it means that team Y wins. This summation of combinations can be also represented as winning 10 matches out of 19 matches.
$\Rightarrow {}^{19}{{C}_{10}}$
Combination is generally given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.$ Expanding this,
$\Rightarrow {}^{19}{{C}_{10}}=\dfrac{19!}{10!\left( 19-10 \right)!}$
The term in the brackets can be subtracted,
$\Rightarrow {}^{19}{{C}_{10}}=\dfrac{19!}{10!9!}$
Now we multiply the numerator and denominator by 20,
$\Rightarrow {}^{19}{{C}_{10}}=\dfrac{19!\times 20}{10!9!\times 20}$
We know that $20\times 19!=20!,$ and we split the 20 in the denominator as $2\times 10,$
$\Rightarrow {}^{19}{{C}_{10}}=\dfrac{20!}{10!9!\times 10\times 2}$
We again know that $10\times 9!=10!,$
$\Rightarrow {}^{19}{{C}_{10}}=\dfrac{20!}{10!10!\times 2}$
The term $\dfrac{20!}{10!10!}$ is nothing but the combination term ${}^{20}{{C}_{10}}.$ Substituting this in the above equation,
$\Rightarrow {}^{19}{{C}_{10}}=\dfrac{{}^{20}{{C}_{10}}}{2}$

Hence, these are the number of ways in which team X wins and we are given that team X wins in $\dfrac{{}^{20}{{C}_{m}}}{2}$ ways are to find m, we compare the two terms and get the value of m as 10.

Note:
We need to know the general form of combinations which is given as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.$ We need to note that here we are adding up all the combinations as the occurrence of one event is independent of the occurrence of the other event.