Question

India and South Africa play one day international series until one team wins 4 matches. No match ends in a draw. Let \[k\] be the number of ways the series can be won. Find the sum of digits of \[k\]?

Answer 1

Hint: First of all, find the number of ways in which the series can win 4 matches by South Africa and then by India. Add them up to find the value of \[k\]. Further sum the digits to get the required answer.

Complete step-by-step answer:
Taking \[I\] for India and \[S\] for South Africa. We can arrange \[I\] and \[S\] to show the wins for India and South Africa respectively.
For example: \[ISSSS\] means the first match is won by India which is followed by 4 wins by South Africa. This is one way in which series can be won.
Suppose South Africa wins the series, then the last match is won by South Africa so for all those cases of different number of matches won by India and South Africa considering the last match won by South Africa we can draw a table as shown below .

	Wins of India	Wins of South Africa	No. of ways
1.	0	4	1
2.	1	4	\[\dfrac{{4!}}{{3!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} = 4\]
3.	2	4	\[\dfrac{{5!}}{{2!3!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2 \times 1} \right)\left( {3 \times 2 \times 1} \right)}} = 10\]
4	3	4	\[\dfrac{{6!}}{{3!3!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {3 \times 2 \times 1} \right)\left( {3 \times 2 \times 1} \right)}} = 20\]

 Therefore, number of ways where South Africa can win 4 matches = 1 + 4 + 10 + 20 = 35
In the same way number of ways where India can win 4 matches = 35
Hence, the total number of ways in which the series can be won = 35+ 35 = 70
Given that the number of ways in which the series can be won is \[k\]. So, \[k = 70\].
Now we have to find the sum of digits in \[k\] i.e., 7 + 0 = 7.
Thus, the sum of digits of \[k\] is 7.

Note: If we have ‘\[n\]’ distant objects of which ‘\[r\]’ objects has to be arranged, then the total number of different arrangements possible are given by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]. Here the permutation doesn’t allow repetition.