Answer:
Two things are to be noted while solving the problem.
1. As Virat has to keep the strike through out the over, he can score only even runs like 2,4,6 or no runs with each ball.
2. As he has to play all the 6 balls of the over , some runs have to be scored from the last ball. He can score the 12 runs from 6 balls in various sequences and the no. of ways of selecting the balls depends accordingly.
1. Only 2 balls in sequence of 6+6 =12:
One ball is already pre selected as last ball. The remaining 1 ball can be selected out of 5 balls in 5 ways.( That is 12 runs will be made from ball 1&6 or 2&6 or 3&6 or 4&6 or 5&6.).So no.of ways for this option = 5
2. Only 3 balls in sequence of 4+4+4= 12.:
Again as one ball is reserved as last ball, we have to select remaining 2 balls out of 5 balls . As per laws of permutations, the selection of " r" objects out of "n" objects can be done in nPr ways= n ! / (n-r)!. If out of the " n" objects , " m" objects are identical then the number of ways of selecting "r" objects is nPr/ m! = n!/ { ( n-r)! * m!}. As Virat has to score identical runs of 4 each from these 2 balls they both constitute 2 identical objects. So number of ways of selecting 2 identical balls out 5 balls is 5P2/ 2!= 5!/{ ( 5-2)! * 2!} = 5!/( 3!*2!). = 10 ways.
3. Only 3 balls in combination of 6+4+2 = 12
Here the last can be hit either for 6, 4, 2 runs. So it can be selected in 3 ways. After that we can select 2 balls out of 5 balls in 5P2 ways. As both the balls will be hit for different amount of runs they are not identical. Therefore the number of ways of selection for the 2 balls is 5P2= 5! /3!= 5*4= 20'ways. As these 20 ways are possible for each of the 3 ways of selecting the last ball , the total number of selection for this option is 3*20 = 60 ways.
4. Only 4 balls in combination of 6+2+2+2 =12:
Here the last ball can be hit either for 2 runs or 6 runs.
Case1: The last ball is hit for 6 runs. Then we have to select 3 balls out of 5 balls in which all the 3 balls will be hit for 2 runs each and hence all 3 are identical. Therefore number of ways of selecting the 3 balls is 5P3 / 3! = 5*4*3/ 6= 10
Case 2: The last ball is hit for 2 runs. Then we have to select 3 balls out of 5 balls in which 2 balls will be hit for 2 runs each and hence they 2 are identical. Therefore the number of ways of doing this is 5P3/2! = 5*4*3/2 = 30. So total no. of ways for this option = 10+30 =40
5. Only 4 number of balls in combination of 4+4+2+2='12.
Here last ball can be selected in 2 ways.It can be hit for 4 or 2 runs.
Case1: Last ball will be hit for 2 runs.From the remaining 5 balls we can select 3 balls out of which 2 balls would be hit for 4 runs each and hence identical..Therefore the number of ways of doing this is 5 P 3/2!=5*4**3/2. = 30
Case 2: Last ball will be hit for 4 runs.
From the remaining 5 balls we can select 3 balls out of which 2 balls would be hit for 2 runs each and hence identical. This situation is same case 1 and so the number of ways of doing this is 5P3/2!= 30. So total number of ways of selection for this option would be 30+ 30 = 60 ways.
6. Only 5 balls in combination of 4+2+2+2+2 = 12.
Here last ball can be hit either for 4 or 2 runs.
Case 1: Last ball will be hit for 4 runs.
From the remaining 5 balls we can select 4 balls in 5 P4 ways and as all the 4 balls will be hit for 2 runs each and so there will be 4 identical balls. Therefore number of ways of selecting 4 balls = 5P4/ 4! = 5!/ ( 1!*4!) = 5 ways.
Case2: Last ball will be hit for 2 runs.
From the remaining 5 balls we can select 4 balls in 5P4 ways.But in this 3 balls will be hit for 2 runs each and so there will be 3 identical balls. Therefore number of ways of selecting 4 balls = 5P4/ 3! = 5!/ ( 1!*3!) = 20 ways. So total number of ways of selection for this option is 5+ 20 = 25 ways
7. All the 6 balls in combination of 2+2+2+2+2+2 =12.
In this option all the 6 balls will be selected at a time and so the number of ways for this option =1
So total number of ways of hitting exactly 12 runs= 5+10+60+40+60+ 25+ 1 = 201
Answer:201
Solution:
Total no. of pairs to score 12 runs in an over:(6,6),(6,4,2),(6,2,2,2),(4,4,4),(4,4,2,2),(4,2,2,2,2),(2,2,2,2,2,2)
No of ways (6,6) can be scored:5C1(One 6 on last ball)
No. of ways (6,4,2) can be scored:3X2X5C2(Any of three(6,4,2) on last ball and remaining two on any of 5 balls)
No. of ways(6,2,2,2) can be scored:5C3+3x5C3
No. of ways(4,4,4) can be scored:5C2
No. of ways(4,4,2,2) can be scored:3X2X5C3
No. of ways(4,2,2,2,2) can be scored:5C4+4X5C4
No. of ways(2,2,2,2,2,2) can be scored:1
Total No. of ways 12 runs can be scored:
5+60+10+30+10+60+5+20+1=201
Where nCr=n!/(n-r)!r!
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