How many different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen, there being one lady and one gentleman on each side.
Answer
680.4k+ views
Hint: First of all, find the number of ways of selecting one lady in each of the two teams and the number of ways of selecting one gentleman in each of the two teams. Then use the multiplicative principle of combinations to get the required answer.
Complete step-by-step answer:
Given number of ladies = 4
Number of gentlemen = 3
Now we have to select 1 gentleman and 1 lady from the total of 3 gentlemen and 4 ladies on each side.
Number of ways of selecting one lady in each of the two teams \[ = {}^4{C_2}\]
Number of ways of selecting one gentleman in each of the two teams \[ = {}^3{C_2}\]
By using multiplicative principle of combinations, the total number of ways of selecting one man and one lady in each of the two teams\[ = {}^4{C_2} \times {}^3{C_2} = \dfrac{{4 \times 3}}{{2 \times 1}} \times \dfrac{{3 \times 2}}{{2 \times 1}} = 6 \times 3 = 18\].
Players of the two teams can interchange between each other, thus the total number of ways \[ = 18 \times 2 = 36\].
Thus, 36 different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen, there being one lady and one gentleman on each side.
Note: In this problem we have used multiplicative principle combinations i.e., if there are \[x\] number of ways of selecting one thing and \[y\] number of ways of selecting another, then the total number of ways of selecting both the things is given in \[xy\] number of ways.
Complete step-by-step answer:
Given number of ladies = 4
Number of gentlemen = 3
Now we have to select 1 gentleman and 1 lady from the total of 3 gentlemen and 4 ladies on each side.
Number of ways of selecting one lady in each of the two teams \[ = {}^4{C_2}\]
Number of ways of selecting one gentleman in each of the two teams \[ = {}^3{C_2}\]
By using multiplicative principle of combinations, the total number of ways of selecting one man and one lady in each of the two teams\[ = {}^4{C_2} \times {}^3{C_2} = \dfrac{{4 \times 3}}{{2 \times 1}} \times \dfrac{{3 \times 2}}{{2 \times 1}} = 6 \times 3 = 18\].
Players of the two teams can interchange between each other, thus the total number of ways \[ = 18 \times 2 = 36\].
Thus, 36 different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen, there being one lady and one gentleman on each side.
Note: In this problem we have used multiplicative principle combinations i.e., if there are \[x\] number of ways of selecting one thing and \[y\] number of ways of selecting another, then the total number of ways of selecting both the things is given in \[xy\] number of ways.
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