Question

Find the number of ways in which 6 persons out of 5 men & 5 women can be seated at a round table such that 2 men are never together.

Answer 1

5400

Answer 2

To find the number of ways to seat 6 persons out of 5 men and 5 women at a round table such that no two men are ever together, we follow these steps:

1. Determine the possible compositions of the 6 persons:
   Let $M_s$ be the number of men selected and $W_s$ be the number of women selected.
   We must have $M_s + W_s = 6$.
   We have a maximum of 5 men and 5 women available.
   For no two men to be together in a circular arrangement, the number of women must be greater than or equal to the number of men ($W_s \ge M_s$).

   Let's list the possible $(M_s, W_s)$ combinations satisfying these conditions:

   • Case 1: 1 Man and 5 Women ($M_s=1, W_s=5$)

     • Condition $W_s \ge M_s$ (5 $\ge$ 1) is satisfied.
     • Selection: $\binom{5}{1}$ ways to choose 1 man from 5, and $\binom{5}{5}$ ways to choose 5 women from 5.
       Number of selections = $\binom{5}{1} \times \binom{5}{5} = 5 \times 1 = 5$.
     • Arrangement: With only 1 man selected, it's impossible for two men to be together. So, any circular arrangement of these 6 people is valid.
       Number of circular arrangements = $(6-1)! = 5! = 120$.
     • Total ways for Case 1 = (Number of selections) $\times$ (Number of arrangements) = $5 \times 120 = 600$.

   • Case 2: 2 Men and 4 Women ($M_s=2, W_s=4$)

     • Condition $W_s \ge M_s$ (4 $\ge$ 2) is satisfied.
     • Selection: $\binom{5}{2}$ ways to choose 2 men from 5, and $\binom{5}{4}$ ways to choose 4 women from 5.
       Number of selections = $\binom{5}{2} \times \binom{5}{4} = 10 \times 5 = 50$.
     • Arrangement: To ensure no two men are together, we use the gap method.
       • First, arrange the 4 women around the table: $(4-1)! = 3! = 6$ ways.
       • These 4 women create 4 distinct spaces between them.
       • Place the 2 selected men into these 4 spaces. Since the men are distinct, this is a permutation.
         Number of ways to place 2 men in 4 spaces = $P(4,2) = \frac{4!}{(4-2)!} = 4 \times 3 = 12$ ways.
       • Total arrangements for selected people = $6 \times 12 = 72$.
     • Total ways for Case 2 = (Number of selections) $\times$ (Number of arrangements) = $50 \times 72 = 3600$.

   • Case 3: 3 Men and 3 Women ($M_s=3, W_s=3$)

     • Condition $W_s \ge M_s$ (3 $\ge$ 3) is satisfied.
     • Selection: $\binom{5}{3}$ ways to choose 3 men from 5, and $\binom{5}{3}$ ways to choose 3 women from 5.
       Number of selections = $\binom{5}{3} \times \binom{5}{3} = 10 \times 10 = 100$.
     • Arrangement:
       • First, arrange the 3 women around the table: $(3-1)! = 2! = 2$ ways.
       • These 3 women create 3 distinct spaces between them.
       • Place the 3 selected men into these 3 spaces.
         Number of ways to place 3 men in 3 spaces = $P(3,3) = 3! = 6$ ways.
       • Total arrangements for selected people = $2 \times 6 = 12$.
     • Total ways for Case 3 = (Number of selections) $\times$ (Number of arrangements) = $100 \times 12 = 1200$.

   • Case 4: 4 Men and 2 Women ($M_s=4, W_s=2$)

     • Condition $W_s \ge M_s$ (2 $\ge$ 4) is NOT satisfied. It's impossible to seat 4 men such that no two are together if there are only 2 women. (If you seat 2 women, there are 2 gaps. Placing 4 men in 2 gaps means at least one gap gets 2 men). So, 0 ways for this case.

   • Case 5: 5 Men and 1 Woman ($M_s=5, W_s=1$)

     • Condition $W_s \ge M_s$ (1 $\ge$ 5) is NOT satisfied. So, 0 ways for this case.

2. Sum the ways from all valid cases:
   Total number of ways = (Ways for Case 1) + (Ways for Case 2) + (Ways for Case 3)
   Total number of ways = $600 + 3600 + 1200 = 5400$.

The final answer is $\boxed{\text{5400}}$.

Explanation of the solution:

1. Identify possible combinations of men and women (M, W) totaling 6 people, considering the available pool (5 men, 5 women) and the condition (no two men together implies W >= M). Valid combinations are (1M, 5W), (2M, 4W), (3M, 3W).
2. For each valid combination: a. Calculate the number of ways to select the men and women using combinations ($\binom{n}{k}$). b. Calculate the number of ways to arrange them at a round table using the gap method: i. Arrange the women circularly: $(W-1)!$. ii. Arrange the men in the spaces created by women using permutations ($P(W,M)$). c. Multiply selections and arrangements for each case.
3. Sum the results from all valid cases.