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5 men and 3 women are to be seated such that no 2 women sit together and 2 men sit together. Find the no of ways in which this can be arranged ?
- a)625
- b)720
- c)525
- d)516
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
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5 men and 3 women are to be seated such that no 2 women sit together a...
Answer – B.720 Explanation : 5!*3! = 120*6 = 720
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5 men and 3 women are to be seated such that no 2 women sit together a...
Problem: To find the number of ways to seat 5 men and 3 women such that no two women sit together and no two men sit together.
Solution:
Step 1: Let us first arrange the 5 men in a row. The number of ways to arrange them is 5!
Step 2: Now, we need to arrange the 3 women. Since no two women can sit together, we need to find the number of ways to arrange them such that they alternate with the men.
- Case 1: If we start with a man, then we need to place the women in the empty spaces between the men. The number of ways to do this is 5C3. Once the women are placed, we can arrange them in 3! ways.
- Case 2: If we start with a woman, then we need to place the men in the empty spaces between the women. The number of ways to do this is 4C2. Once the men are placed, we can arrange them in 2! ways.
Step 3: Since we can start with either a man or a woman, we need to add the results of Case 1 and Case 2.
Total number of ways = 5! x [(5C3 x 3!) + (4C2 x 2!)]
= 120 x [(10 x 6) + (6 x 2)]
= 7200
Therefore, the number of ways to seat 5 men and 3 women such that no two women sit together and no two men sit together is 720.
Answer: Option B.
Solution:
Step 1: Let us first arrange the 5 men in a row. The number of ways to arrange them is 5!
Step 2: Now, we need to arrange the 3 women. Since no two women can sit together, we need to find the number of ways to arrange them such that they alternate with the men.
- Case 1: If we start with a man, then we need to place the women in the empty spaces between the men. The number of ways to do this is 5C3. Once the women are placed, we can arrange them in 3! ways.
- Case 2: If we start with a woman, then we need to place the men in the empty spaces between the women. The number of ways to do this is 4C2. Once the men are placed, we can arrange them in 2! ways.
Step 3: Since we can start with either a man or a woman, we need to add the results of Case 1 and Case 2.
Total number of ways = 5! x [(5C3 x 3!) + (4C2 x 2!)]
= 120 x [(10 x 6) + (6 x 2)]
= 7200
Therefore, the number of ways to seat 5 men and 3 women such that no two women sit together and no two men sit together is 720.
Answer: Option B.
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