Question

A batsman scores exactly a century by hitting fours and sixes in 20 consecutive balls. In how many ways can he do it if some balls may not yield runs and the order of boundaries and over boundaries are taken into account?

Answer 1

Hint: In this question the concept of permutations and combinations will be used. We need to apply logic and general algebra to proceed through the problem.We will assume three variables x, y and z for the number of fours, sixes and zero runs respectively. Then we will form an equation that the sum of balls is 20. The second equation will be the sum of runs scored through fours, sixes and zeros should be 100. The number of ways to arrange n things when r things are identical are-
$\dfrac{{{\text{n}}!}}{{{\text{r}}!}}$

Complete step-by-step answer:
The batsman can hit a four, a six or zero runs at every ball. Let the number of fours be x, the number of sixes be y and the zeros be z. There are a total of 20 balls so-
${\text{x}} + {\text{y}} + {\text{z}} = 20,\;0 \leqslant {\text{x}},\;{\text{y}},\;{\text{z}} \leqslant 20$
x = 20 - y - z …(1)

Also, it is given that the total number of runs scored are 100 so-
$4{\text{x}} + 6{\text{y}} + 0{\text{z}} = 100$
We will substitute the equation (1) in this equation-
$4\left( {20 - {\text{y}} - {\text{z}}} \right) + 6{\text{y}} = 100$
$80 - 4{\text{y}} - 4{\text{z}} + 6{\text{y}} = 100$
$2{\text{y}} - 4{\text{z}} = 20$
${\text{y}} = 10 + 2{\text{z}}$
Now, we will substitute the values of z to find the corresponding values of x and y as-
When z = 0,
y = 10 + 2(0) = 10
x = 20 - 10 - 0 = 10
The solution (x, y, z) = (10, 10, 0)
The number of ways to score runs in this way are-
$\dfrac{{20!}}{{10!10!0!}}$
When z = 1,
y = 10 + 2(1) = 12
x = 20 - 12 - 1 = 7
The solution (x, y, z) = (7, 12, 1)
The number of ways to score runs in this way are-
$\dfrac{{20!}}{{7!12!1!}}$

When z = 2,
y = 10 + 2(2) = 14
x = 20 - 14 - 2 = 4
The solution (x, y, z) = (4, 14, 2)
The number of ways to score runs in this way are-
$\dfrac{{20!}}{{4!14!2!}}$

When z = 3,
y = 10 + 2(3) = 16
x = 20 - 16 - 3 = 1
The solution (x, y, z) = (1, 16, 3)
The number of ways to score runs in this way are-
$\dfrac{{20!}}{{1!16!3!}}$

When z = 4,
y = 10 + 2(4) = 18
x = 20 - 18 - 4 = -2
x cannot be negative hence this case is not possible.
 Hence the total number of ways are-
$\dfrac{{20!}}{{10!10!0!}} + \dfrac{{20!}}{{7!12!1!}} + \dfrac{{20!}}{{4!14!2!}} + \dfrac{{20!}}{{1!16!3!}}$

This is the required number of ways.


Note: There is no direct method to solve this problem. We need to form the equations according to the question, and simplify it using general algebra. One common mistake is that students often forget the formula for combinations and permutations, which should be remembered. Another mistake is that students use the values of z which do not satisfy the constraints. For example, they can use z = 4, without considering the fact that the corresponding value of x is a negative number, which is not possible.