Question 9 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Class X1 - Maths -Permutations and Combinations Page 157
Answers
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There are 4 even place , so 4 women want to sit.
hence, number of ways can be done =4!
Now, remaining 5 seats can be filled be men.
number of ways can be filled by men=5!
Hence, by Fundamental principle of counting,
Total number of ways = 5! × 4!
= {5 × 4 × 3 × 2}×{4 × 3 × 2}
= 120 × 24
= 2880 ways .
Complete step-by-step answer:
First, before we solve this problem, we try to understand how we can arrange 5 men and 4 women in a row without any constraints. To count the number of combinations for this case, we basically have 9 people (we have no constraints so there is no difference between men and women while counting the number of combinations), we have the combinations as 9!
However, the required number of cases with constraints, clearly we would have a lesser number of cases than 9! arrangements. Thus, to give an idea about the current problem, we have,
M W M W M W M W M (Basically, our arrangement would look like this since women would occupy even places.)
Thus, the places for the women are fixed, thus the total number of arrangements is 4! (which is 4P4).
Similarly, due to the women’s places getting fixed, men’s places automatically fixed, thus the total number of arrangements is 5! (which is 5P5). Now, combining these results, we have a total number of combinations of 4!×5!=2880.