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In how many ways can 3 teachers and 4 pupils be arranged in a line if the pupils and teachers must alternate? . how to get the answer? the ans :144

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  • $\begingroup$ You know which seats are for children and which for teachers, so just arrange the children in the children's seats and the teachers in the teachers' seats. You might get more helpful answers if you showed your own thoughts on your questions $\endgroup$
    – Henry
    Commented Sep 10, 2014 at 11:35

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Let '|'-represent students and '*'-represent the gaps between them where you need to place the teachers for alternate configuration then:

            | * | * | * | 

So now you can permute the 4 students in 4! ways and on placing 3 teachers in the gap you can permute them in 3! ways

Therefore, No. of ways of arranging them in alternate position=3!*4!=6*24=144,which is your answer

Since it is 'and' segment so we are multiplying( 3 teachers and 4 pupils)

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Hint: first, show that the pupils must be on the first, third, fifth and seventh places. Then, think about how many ways you can place the pupils on those $4$ places and how many ways you can place the teachers on the remaining. Are the two arrangements dependent on each other?

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