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In how many ways can 4 men and 3 women be arranged in a round table:  i) if the women always sit together? ii) if the women never sit together?

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For case $1$: First fix the women in one group. $3$ women can be seated in $3!$ ways. The $4$ men can be seated in the remaining $4$ places in $4!$ ways. Therefore, there are $144$ arrangements.

For case $2$: First fix the men in circular arrangement, which can be done in $3!$ following from the formula $(n-1)! $. Now there are $4$ places between the men and three women are to be seated between the men. That can be done in $^4 P_3$ ways. Again you get $144$ arrangements.

Hope it helps.

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If the women have to sit together, they from a group which can be arranged in $3!$ possibilities. Mixing this group with the 4 men, we have a total of $3!5!$ options.

If the women ought not to sit together, it implies there must be one guy next to them (think about why it cannot be two). This gives us: $3!4!$ options.

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Case 1 - All women sit together.

Take women as 1 group. So we now have 4 men and 1 group . These 5 men can arranged in round table in (5-1)! = 4! = 24 ways.

But women can be further arranged in group in 3!= 6 ways.

Therefore total number of ways is equal to 24 × 6 = 144 ways .

Case 2- No women sit together.

As number of women is less than number of men , two of the men will be in between a pair of women. So fix the seats of women.

Women can be arranged in 3!= 6 ways.

Now select two men. They can be selected in C(4,2) = 6 ways and arranged in 2! = 2 ways.

Consider these 2 men as a single man. Now we have 3 men and they can be arranged in the round table in (3-1)! = 2! = 2 ways .

Therefore total number of ways = 6 × 6 × 2 × 2 = 144 ways .

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