Question

Find the number of ways in which five bowlers of different ages can choose to bowl one over each one after the other, if the oldest brother should not bowl the first over.
A. 96
B. 84
C. 92
D. 98

Answer 1

Here we use the concept of combinations to find the number of ways for an over to be bowled by a bowler. Since we have to keep exception for the first over that cannot have the oldest bowler, we reduce one bowler from the total number of bowlers and then choose.

• Combination formula is given by $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, where n is total number of items w are choosing from and r is number of items we are selecting.

Complete step-by-step answer:
Let us assume the five bowlers as ${B_1},{B_2},{B_3},{B_4},{B_5}$ where the bowler ${B_5}$is the oldest.
Now we find a number of ways an over can be bowled separately.
First over:
We have total number of bowlers as 5
But since first over cannot be bowled by the oldest bowler, so we have number of bowlers as $5 - 1 = 4$
So, $n = 4$
Now only one bowler can bowl at a time, so we take $r = 1$.
Therefore, the number of ways one player can bowl first over is $^4{C_1}$.
Solve the value of$^4{C_1}$ using the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
${ \Rightarrow ^4}{C_1} = \dfrac{{4!}}{{(4 - 1)!1!}}$
${ \Rightarrow ^4}{C_1} = \dfrac{{4!}}{{3!1!}}$
We know factorial opens up as $n! = n(n - 1)!$
${ \Rightarrow ^4}{C_1} = \dfrac{{4 \times 3!}}{{3!1!}}$
Cancel the same terms from numerator and denominator and put $1! = 1$
${ \Rightarrow ^4}{C_1} = 4$
Second over:
Now we have a total number of bowlers as 4.
So, $n = 4$
Now only one bowler can bowl at a time, so we take $r = 1$.
Therefore, the number of ways one player can bowl second over is $^4{C_1}$.
Solve the value of$^4{C_1}$ using the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
${ \Rightarrow ^4}{C_1} = \dfrac{{4!}}{{(4 - 1)!1!}}$
${ \Rightarrow ^4}{C_1} = \dfrac{{4!}}{{3!1!}}$
We know factorial opens up as $n! = n(n - 1)!$
${ \Rightarrow ^4}{C_1} = \dfrac{{4 \times 3!}}{{3!1!}}$
Cancel the same terms from numerator and denominator and put $1! = 1$
${ \Rightarrow ^4}{C_1} = 4$
Third over:
Now we have a total number of bowlers as 3.
So, $n = 3$
Now only one bowler can bowl at a time, so we take $r = 1$.
Therefore, the number of ways one player can bowl third over is $^3{C_1}$.
Solve the value of$^3{C_1}$ using the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
${ \Rightarrow ^3}{C_1} = \dfrac{{3!}}{{(3 - 1)!1!}}$
${ \Rightarrow ^3}{C_1} = \dfrac{{3!}}{{2!1!}}$
We know factorial opens up as $n! = n(n - 1)!$
${ \Rightarrow ^3}{C_1} = \dfrac{{3 \times 2!}}{{2!1!}}$
Cancel the same terms from numerator and denominator and put $1! = 1$
${ \Rightarrow ^3}{C_1} = 3$
Fourth over:
Now we have a total number of bowlers as 2.
So, $n = 2$
Now only one bowler can bowl at a time, so we take $r = 1$.
Therefore, the number of ways one player can bowl fourth over is $^2{C_1}$.
Solve the value of$^2{C_1}$ using the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{(2 - 1)!1!}}$
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{1!1!}}$
We know factorial opens up as $n! = n(n - 1)!$
${ \Rightarrow ^2}{C_1} = \dfrac{{2 \times 1!}}{{1!1!}}$
Cancel the same terms from numerator and denominator and put $1! = 1$
${ \Rightarrow ^2}{C_1} = 2$
Fifth over:
Now we have a total number of bowlers as 1.
So, $n = 1$
Now only one bowler can bowl at a time, so we take $r = 1$.
Therefore, the number of ways one player can bowl fifth over is $^1{C_1}$.
Solve the value of$^1{C_1}$ using the formula $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
${ \Rightarrow ^1}{C_1} = \dfrac{{1!}}{{(1 - 1)!1!}}$
${ \Rightarrow ^1}{C_1} = \dfrac{{1!}}{{0!1!}}$
Cancel the same terms from numerator and denominator and put $0! = 1$
${ \Rightarrow ^1}{C_1} = 1$
Therefore, the total number of ways in which 5 bowlers can bowl is given by multiplication of the number of ways each over can be bowled.
Number of ways to bowl $= 4 \times 4 \times 3 \times 2 \times 1$
Number of ways to bowl $= 96$

So, the correct option is A.

Note: Alternate method:
Students can solve the number of ways to bowl 5 overs without the use of combinations also.
Since there are 5 overs to bowl and 5 players
Choices of bowler for first over is 4 (because the oldest bowler is not included)
Choices of bowler for second over is 4 (because we have 4 bowlers left)
Choices of bowler for third over is 3 (because we have 3 bowlers left)
Choices of bowler for fourth over is 4 (because we have 2 bowlers left)
Choices of bowler for fifth over is 4 (because we have 1 bowlers left)
So, total number of ways to bowl $= 4 \times 4 \times 3 \times 2 \times 1$
Number of ways to bowl $= 96$