Question

A batsman can score 0, 1, 2, 3, 4, or 6 runs from a ball. The number of different sequences in which he can score exactly 30 runs in an over of six balls, is
(a) 4
(b) 71
(c) 56
(d) 76

Answer 1

Hint: We start solving the problem by writing all possibilities of getting 30 runs in an over. We then find the total sequences that can be formed for each possibility using the result that the total number of ways of arranging the ‘n’ objects in which ‘r’ of one type, ‘s’ of one type, ‘t’ of one type is given by $\dfrac{n!}{r!s!t!}\left( n > r,s,t \right)$. We then add the number of sequences that were obtained for the possibilities to find the total number of sequences satisfying the given condition in the problem.

Complete step-by-step solution:
According to the problem, we need to find the total number of different sequences in which the batsman can score exactly 30 runs in an over of six balls if a batsman can score 0, 1, 2, 3, 4, or 6 runs from a ball.
Let us first write what are the possible ways to get 30 runs in an over of 6 balls.
The possibilities of getting 30 runs are as shown below (need to specific in that order):
(i) 6, 6, 6, 6, 6, 0
(ii) 6, 6, 6, 6, 4, 2
(iii) 6, 6, 6, 4, 4, 4
(iv) 6, 6, 6, 6, 3, 3
Let us find the total number of distinct sequences for each possibility.
We know that the total number of ways of arranging the ‘n’ objects in which ‘r’ of one type, ‘s’ of one type, ‘t’ of one type is given by $\dfrac{n!}{r!s!t!}\left( n > r,s,t \right)$. We also know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 3\times 2\times 1$.
So, the total number of different sequences occurring for the possibility (i) is $\dfrac{6!}{5!}=6$.
The total number of different sequences occurring for the possibility (ii) is $\dfrac{6!}{4!}=6\times 5=30$.
The total number of different sequences occurred for the possibility (iii) is $\dfrac{6!}{3!.3!}=\dfrac{6\times 5\times 4}{3\times 2\times 1}=20$. The total number of different sequences occurring for the possibility (iv) is $\dfrac{6!}{4!.2!}=\dfrac{6\times 5}{2\times 1}=15$.
We know that these possibilities are independent of each other. So, we need to add these numbers of sequences for each possibility.
So, the total number of sequences that can be formed is $6+30+20+15=71$.
So, we have found that the number of different sequences in which he can score exactly 30 runs in an over of six balls is 71.
The correct option for the given problem is (b).

Note: Whenever we get this type of problem, we first try to write the possible ways of getting the solution as it provides half of the answer to us. We should only use the scores given in the problem for each ball which is very important while solving this problem. We should not confuse it with a multiplication of numbers while finding the factorial. Similarly, we can expect problems to find the total number of sequences if a batsman can score 1, 2, 3, 4, 5, or 6 from a ball.