Question

The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no husband and wife play in the same game, is

A. 46
B. 54
C. 60  
D. None of these
Answer :   60
Solution :
Let the sides of the game be $$A$$ and $$B.$$ Given 5 married couples, i.e., 5 husbands and 5 wives. Now, 2 husbands for two sides $$A$$ and $$B$$ can be selected out of 5 $$ = {\,^5}{C_2} = 10{\text{ ways}}{\text{.}}$$
After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game). So we have to choose 2 wives out of remaining $$5 - 2 = 3$$   wives i.e., $$ = {\,^3}{C_2} = 3{\text{ ways}}{\text{.}}$$
Again two wives can interchange their sides $$A$$ and $$B$$ in $$2! = 2{\text{ ways}}{\text{.}}$$
By the principle of multiplication, the required number of ways $$ = 10 \times 3 \times 2 = 60$$

Releted MCQ Question on
Algebra >> Permutation and Combination

Releted Question 1

$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$     and $$^n{C_{r + 1}} = 126,$$   then $$r$$ is:

A. 1
B. 2
C. 3
D. None of these.
Releted Question 2

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are

A. 69760
B. 30240
C. 99748
D. none of these
Releted Question 3

The value of the expression $$^{47}{C_4} + \sum\limits_{j = 1}^5 {^{52 - j}{C_3}} $$    is equal to

A. $$^{47}{C_5}$$
B. $$^{52}{C_5}$$
C. $$^{52}{C_4}$$
D. none of these
Releted Question 4

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

A. $$^6{C_3} \times {\,^4}{C_2}$$
B. $$^4{P_2} \times {\,^4}{C_3}$$
C. $$^4{C_2} + {\,^4}{P_3}$$
D. none of these

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Permutation and Combination


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