Question

In a unique hockey series between India and Pakistan, they decide to play on till a team wins $5$ matches. The number of ways in which the series can we won by India, if no match ends in draw is:
  (A)$126$
  (B)$252$
  (C)$225$
  (D)None of these

Answer 1

Hint: In these questions to calculate the number of ways in any situation we use the method of permutation and combination. In this question we will use the formula of combination which is given by
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. With this formula this question can be easily solved.

Complete step-by-step answer: To win series, India have to win $5$matches, this can be done in five different ways which are
$\left( 1 \right)$ India will win $5$ consecutive matches.
$\left( 2 \right)$ India will win $4$ matches and $1$ loss , Hence for the winning of India we consider the last match will be won by India hence total matches will be $6$.
$\left( 3 \right)$ India will win $4$ matches and $2$ loss , Hence for the winning of India we consider the last match will be won by India hence total matches will be $7$.
$\left( 4 \right)$ India will win $4$ matches and $3$ loss , Hence for the winning of India we consider the last match will be won by India hence total matches will be $8$.
$\left( 5 \right)$ India will win $4$ matches and $4$ loss , Hence for the winning of India we consider the last match will be won by India hence total matches will be $9$.
As the above we did not consider the fifth winning match because as India will win the last match the event will stop at that point .
So we have to apply Combination only in the initial four matches .

Now we have to find the number of total ways in which India wins in each way given above.
Now by using a combination formula.
$ \Rightarrow $ India will win $5$ consecutive matches$ = 1$
$ \Rightarrow $ India will win $4$ matches and $1$ loss $ = {}^5{C_4}$
$ \Rightarrow $ India will win $4$ matches and $2$ loss $ = {}^6{C_4}$
$ \Rightarrow $ India will win $4$ matches and $3$ loss $ = {}^7{C_4}$
$ \Rightarrow $ India will win $4$ matches and $4$ loss $ = {}^8{C_4}$
$\therefore $ Total number of ways $ = $$1 + {}^5{C_4} + {}^6{C_4} + {}^7{C_4} + {}^8{C_4}$
$ \Rightarrow $$1 + \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} + \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}} + \dfrac{{7!}}{{4!\left( {7 - 4} \right)!}} + \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}}$
$ = $$1 + \dfrac{{5!}}{{4!1!}} + \dfrac{{6!}}{{4!2!}} + \dfrac{{7!}}{{4!3!}} + \dfrac{{8!}}{{4!4!}}$
Now Canceling out the common term from numerator and denominator hence the remaining equation become ,
$ = $$1 + \dfrac{5}{1} + \dfrac{{6 \times 5}}{2} + \dfrac{{7 \times 6 \times 5}}{{3 \times 2}} + \dfrac{{8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2}}$
Hence after solving more we get ,
\[1 + 5 + \dfrac{{3 \times 5}}{1} + \dfrac{{7 \times 5}}{1} + \dfrac{{7 \times 2 \times 5}}{1}\]
\[1 + 5 + 15 + 35 + 70\]
$ = 126$
Hence, the total number of ways is $126$

So, the correct answer is “Option A”.

Note: To apply the combination formula, one has to understand all the situations which can occur according to given conditions in question. The common mistake that one should make by taking a lesser amount of situations, which can possibly give wrong answers.