four men and three women are to be seated for a dinner such that no two women sit together and no two men sit together. find the number of ways in which this can be arranged.
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Answered by
14
answer is 4! 3! = 24*6 = 144
4! for arranging 4 man in a row first
like
M1 M2 M3 M4
NOW WE can see three gaps b/w men
so we will place 3 women in those 3 gaps by 3!
thus answer 4!*3! = 144 ways
4! for arranging 4 man in a row first
like
M1 M2 M3 M4
NOW WE can see three gaps b/w men
so we will place 3 women in those 3 gaps by 3!
thus answer 4!*3! = 144 ways
Answered by
2
The arrangement will be as follows:-
M , W , M , W , M , W , M
And if the first man and last man are to be seemed together, then place one more chair between them.
Please mark this as the brainliest answer.
GOOD LUCK....
M , W , M , W , M , W , M
And if the first man and last man are to be seemed together, then place one more chair between them.
Please mark this as the brainliest answer.
GOOD LUCK....
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