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There are three men have 3 coats,4 waist coats and 6 caps. In how many ways can they wear them? Also ensure that no 3 men should wear both caps and coats?

 

Option: 1

12,560


Option: 2

64,640


Option: 3

86,420


Option: 4

92,160


Answers (1)

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To find the number of ways the three men can wear the coats, waistcoats, and caps while ensuring that no three men wear both caps and coats, we can use permutations.

For the first man, there are 3 choices for the coat, 4 choices for the waistcoat, and 6 choices for the cap. Therefore, the number of ways the first man can wear the items is
\mathrm{P(3,1) \times P(4,1) \times P(6,1)=3 \times 4 \times 6=72}.

For the second man, since he cannot wear both a coat and a cap, there are 2 choices for the coat (excluding the one already chosen by the first man), 4 choices for the waistcoat, and 5 choices for the cap (excluding the one already chosen by the first man). Therefore, the number of ways the second man can wear the items is .

\mathrm{P(2,1) \times P(4,1) \times P(5,1)=2 \times 4 \times 5=40}.

For the third man, there is only 1 choice left for the coat (excluding the ones already chosen by the first and second men), 4 choices for the waistcoat, and 4 choices for the cap (excluding the ones already chosen by the first and second men). Therefore, the number of ways the third man can wear the items is .

\mathrm{P(1,1) \times P(4,1) \times P(4,1)=1 \times 4 \times 4=16}

To find the total number of ways they can wear all the items, we multiply the number of options for each man together:
\mathrm{P(3,1) \times P(4,1) \times P(6,1) \times P(2,1) \times P(4,1) \times P(5,1) \times P(1,1) \times P(4,1) \times P(4,1)=72 \times 40 \times 16=92,160 . }

Therefore, there are 92,160 different ways the three men can wear the coats, waistcoats, and caps, ensuring that no three men wear both caps and coats.

 

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