Combination

EXERCISE 17.2

QUESTION 1

From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can
this be done?

Sol :

Required number of ways =~^{15}C_{11}

\mathrm{Now}_{,}~^{15}C_{11}=~^{15}C_{4} \left[\because~^{n}C_{r}=~^{n}C_{n-r}\right]

=\dfrac{15}{4} \times \dfrac{14}{3} \times \dfrac{13}{2} \times \dfrac{12}{1} \times~^{11} C_{0} \left[\because~^{n}C_{r}=\dfrac{n}{r}~^{n-1}C_{r-1}\right]

= 1365

 

QUESTION 2

How many different boat parties of 8 , consisting of 5 boys and 3 girls, can be made from 25 boys
and 10 girls?

Sol :

Clearly, out of the 25 boys and 10 girls, 5 boys and 3 girls will be chosen.

Then, different boat parties of 8=~^{25} C_{5} \times~^{10} C_{3}

=\dfrac{25 !}{5 ! 20 !} \times \dfrac{10 !}{3 ! 7 !}

=\dfrac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1} \times \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1}

= 6375600

 

QUESTION 3

In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory
for every student?

Sol :

We are given that 2 courses are compulsory out of the 9 available courses,

Thus, a student can choose 3 courses out of the remaining 7 courses.

Number of ways =7 C_{3}

=\dfrac{7 !}{3 ! 4 !}

=\dfrac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}

= 35

 

QUESTION 4

In how many ways can a football team of 11 players be selected from 16 players? How many of
these will

(i) include 2 particular players?

(ii) exclude 2 particular players?

Sol :

Number of ways in which 11 players can be selected out of 16

^{16} C_{11}=\dfrac{16 !}{11 ! 5 !}

=\dfrac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}

= 4368

(i) If 2 particular players are included, it would mean that out of 14 players, 9 players are
selected.

Required number of ways =14 C_{9}=\dfrac{14 !}{9 ! 5 !} 

=\dfrac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1}

= 2002

(ii) If 2 particular players are excluded, it would mean that out of 14 players, 11 players are
selected.

Required number of ways =~^{14} C_{11}=\dfrac{14 !}{11 ! 3 !}

=\dfrac{14 \times 13 \times 12}{3 \times 2 \times 1}

= 364

 

QUESTION 5

There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students
is to be formed. Find the number of ways in which this can be done. Further find in how many of
these committees: 

(i) a particular professor is included.

(ii) a particular student is included.

(iii) a particular student is excluded.

Sol :

Clearly, 2 professors and 3 students are selected out of 10 professors and 20 students ,respectively.

Required number of ways =~^{10} C_{2} \times~^{20} C_{3}

=\dfrac{10}{2} \times \dfrac{9}{1} \times \dfrac{20}{3} \times \dfrac{19}{2} \times \dfrac{18}{1}

= 51300

(i) If a particular professor is included, it means that 1 professor is selected out of the remaining 9
professors.

Required number of ways =~^{20} C_{3} \times~^{9} C_{1}

=\dfrac{20}{3} \times \dfrac{19}{2} \times \dfrac{18}{1} \times \dfrac{9}{1}

= 10260

(ii) If a particular student is included, it means that 2 students are selected out of the remaining 19
students.

Required number of ways =~^{19} C_{2} \times~^{10} C_{2}

=\dfrac{19}{2} \times \dfrac{18}{1} \times \dfrac{10}{2} \times \dfrac{9}{1}

= 7695

(iii) If a particular student is excluded, it means that 3 students are selected out of the remaining
19 students.

Required number of ways =~^{19}C_{3} \times~^{10} C_{2}

=\dfrac{19}{3} \times \dfrac{18}{2} \times \dfrac{17}{1} \times \dfrac{10}{2} \times \dfrac{9}{1}

= 43605

 

QUESTION 6

How many different products can be obtained by multiplying two or more of the numbers 3 , 5 , 7
, 11 (without repetition)?

Sol :

Required number of ways of getting different products 

=~^{4} C_{2}+~^{4} C_{3}+~^{4} C_{4}

=~^{4} C_{2}+~^{4} C_{3}+1 \left[\because~^nC_n=1\right]

=~^{4} C_{2}+~^{4} C_{4-3}+1  \left[\because~^nC_r=~^nC_{n-r}\right]

=\dfrac{4}{2}\times ^3C_{1}+1 \left[\because~^nC_r=\dfrac{n}{r}~^{n-1}C_{r-1}\right]

=6+4+1=11 \left[\because~^nC_1=n\right]

 

QUESTION 7

From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition; at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?

Sol :

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