Question

In an exciting finish to a one-day cricket match between India and Australia, the Indians require $10$ runs in the last $3$ balls to win. If any one of the scores $0, 1, 2, 3, 4, 5, 6$ can be made from a ball and no wides or no balls are bowled then in how many different sequences can the batsmen make exactly $10$ runs?
(A) $14$
(B) $21$
(C) $36$
(D) $28$

Answer 1

Hint:
For making $10$ runs in $3$ balls is possible if the score of each ball has a sum of $10$. You can make a sequence by taking three numbers from $0, 1, 2, 3, 4, 5, 6$ such that the sum of all three numbers should be ten. This can be done by making different triplets and counting the total number of arrangements.

Complete step by step solution:
Here in this problem, we are given a scenario where $10$ runs are required in the last $3$ balls to win a cricket match. The possible scores on a ball are $0,1,2,3,4,5,6$, with the condition of no wides or no balls. With this information we need to find the number of possible sequences where the batsmen can score exactly $10$ runs in three balls.
To find the number of sequences, we need to find three numbers from $0,1,2,3,4,5,6$ with a sum of $10$.
Therefore, we have:
Sequences with three different score are:
$ \Rightarrow 2 + 3 + 5 = 10$ , which gives us $3! = 6$ ways of arranging $3$ numbers (scores) on $3$ different places (balls)
$ \Rightarrow 6 + 3 + 1 = 10$ , this also gives us $3! = 6$ ways of arranging $3$ numbers (scores) on $3$ different places (balls)
$ \Rightarrow 6 + 4 + 0 = 10$ , which gives us $3! = 6$ ways of arranging $3$ numbers (scores) on $3$ different places (balls)
$ \Rightarrow 5 + 4 + 1 = 10$ , which gives us $3! = 6$ ways of arranging $3$ numbers (scores) on $3$ different places (balls)
Sequences with two similar scoring balls are:
$ \Rightarrow 6 + 2 + 2 = 10$ , which gives us $3$ ways of arranging score six at three places (balls)
$ \Rightarrow 4 + 3 + 3 = 10$ , which gives us $3$ ways of arranging score four at three places (balls)
$ \Rightarrow 5 + 5 + 0 = 10$ , which gives us $3$ ways of arranging score zero at three places (balls)
$ \Rightarrow 4 + 4 + 2 = 10$ , which gives us $3$ ways of arranging score two at three places (balls)
So, we have the total number of possible sequences as: $6 + 6 + 6 + 6 + 3 + 3 + 3 + 3 = 24 + 12 = 36$
Therefore, we get that there are $36$ possible sequences where $10$ runs are made in $3$ balls

Hence, the option (C) is the correct answer.

Note:
In questions like this the utilization of properties from the arrangement is very useful. For calculating the number of arrangements of ‘m’ number of things on ‘m’ places, then we find $m!$ and when ‘r’ things r same among ‘m’ things, the arrangements can be calculated by: $\dfrac{{m!}}{{r!}}$ . We used this property to calculate the number of sequences of the score.