Question

In a group of 13 cricket players four are bowlers. Find out in how many ways can they form a cricket team of 11 players in which at least 2 bowlers are included
A. 55
B. 72
C. 78
D. None of these

Answer 1

We use the method of combination to form a team of 11 players from total 13 players. Form three cases with 2 bowlers, three bowlers and 4 bowlers and add the three possible ways. Write the number of bowlers and non-bowlers and make the formula of combination according to each case.

• 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.
• Factorial terms open up as $n! = n(n - 1)!$

Complete step-by-step solution:
We are given total number of players is 13
Number of bowlers from 13 players $= 4$
Number of non-bowlers from 13 players$= 13 - 4$
$\Rightarrow$Number of non-bowlers from 13 players$= 9$
Now we have to choose 11 players from a total 13 players such that there are at least 2 bowlers. Since we have at least 2 bowlers then we form 3 cases:
First case: 2 bowlers
We have to find a number of ways to make a team of 11 players where 2 bowlers are present and 9 non-bowlers are present.
So we choose 2 bowlers from total 4 bowlers and 9 non-bowlers from 9 non-bowlers.
$\Rightarrow$Number of ways to choose a team ${ = ^4}{C_2}{ \times ^9}{C_9}$
Use the formula of combination to open up the value
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{(4 - 2)!2!}} \times \dfrac{{9!}}{{(9 - 9)!9!}}$
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{2!2!}} \times \dfrac{{9!}}{{0!9!}}$
Open the terms with help of factorial
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4 \times 3 \times 2!}}{{2!2 \times 1}} \times \dfrac{{9!}}{{0!9!}}$
Cancel same factors from numerator and denominator of fractions
$\Rightarrow$Number of ways to choose a team $= 6$..............… (1)
Second case: 3 bowlers
We have to find a number of ways to make a team of 11 players where 3 bowlers are present and 8 non-bowlers are present.
So we choose 3 bowlers from total 4 bowlers and 9 non-bowlers from 9 non-bowlers.
$\Rightarrow$Number of ways to choose a team ${ = ^4}{C_3}{ \times ^9}{C_8}$
Use the formula of combination to open up the value
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{(4 - 3)!3!}} \times \dfrac{{9!}}{{(9 - 8)!8!}}$
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{1!3!}} \times \dfrac{{9!}}{{1!8!}}$
Open the terms with help of factorial
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4 \times 3!}}{{3! \times 1}} \times \dfrac{{9 \times 8!}}{{1!8!}}$
Cancel same factors from numerator and denominator of fractions
$\Rightarrow$Number of ways to choose a team $= 4 \times 9$
$\Rightarrow$Number of ways to choose a team $= 36$...............… (2)
Third case: 4 bowlers
We have to find a number of ways to make a team of 11 players where 4 bowlers are present and 7 non-bowlers are present.
So we choose 4 bowlers from total 4 bowlers and 7 non-bowlers from 9 non-bowlers.
$\Rightarrow$Number of ways to choose a team ${ = ^4}{C_4}{ \times ^9}{C_7}$
Use the formula of combination to open up the value
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{(4 - 4)!4!}} \times \dfrac{{9!}}{{(9 - 7)!7!}}$
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{0!4!}} \times \dfrac{{9!}}{{2!7!}}$
Open the terms with help of factorial
$\Rightarrow$Number of ways to choose a team $= \dfrac{{4!}}{{4! \times 0!}} \times \dfrac{{9 \times 8 \times 7!}}{{2!7!}}$
Cancel same factors from numerator and denominator of fractions
$\Rightarrow$Number of ways to choose a team $= 9 \times 4$
$\Rightarrow$Number of ways to choose a team $= 36$.................… (3)
Total number of ways to make a team is given by sum of values from equations (1), (2) and (3)
$\Rightarrow$Number of ways to form a team of 11 players $= 6 + 36 + 36$
$\Rightarrow$Number of ways to form a team of 11 players $= 78$

$\therefore$Correct option is C.

Note: Many students make the mistake of assuming the value of $0!$ as 0, keep in mind the value of $0! = 1$ is fixed and is not equal to 0. Also, when taking at least cases we take the possibilities or cases with that case and all other higher possibilities as well.