Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In how many ways can you arrange $4$ men and $4$ women in a row of $8$ seats if one man and a woman will insist not to be seated together?

share|improve this question

2 Answers 2

We have $4$ men and $4$ women, and we cannot have $2$ members of the same gender sitting next to each other. So, for the first seat, we have $8$ possible choices. The second seat has $4$ possible choices (the $4$ corresponding to the opposite gender from the individual chosen in the first seat); the third and fourth seats both have $3$ possible choices; the fifth and sixth seats both have $2$ possible choices; and one choice exists for both the seventh and eighth seat. Hence, we have $8.4.3.3.2.2.1.1=1152$ different ways to alternate the males and females.

share|improve this answer
    
hi thanks for the quick response :D –  Amelia Amore Sep 11 '13 at 9:21
    
@Amelia If the answer is correct, you can accept the answer, and if it is not, what is the correct answer?, because i think this is homework –  Ramanujan Sep 11 '13 at 9:33
    
thanks for reminding me, ill come back again tomorrow after we discuss this topic. :D –  Amelia Amore Sep 11 '13 at 9:41
1  
I don't see anything in the statement of the question that says no 2 people of the same gender can sit next to each other. It seems to me that what the question says is that there is one particular man and one particular woman who refuse to sit next to each other. This is also the interpretation @Harish makes when giving what appears to me to be the correct answer. –  Gerry Myerson Sep 27 '13 at 23:21
    
@GerryMyerson Thanks a lot. –  Ramanujan Sep 28 '13 at 3:25

HINT : Arrange 3 men and 3 women (assuming these six people don't mind sitting anywhere) in $6!$ ways and in the 7 places got after arranging six people, choose 2 places and place the 2 people who insist not being together in $7 \choose 2$ * $2! $ ways

answer would be $6!$ $\times$ $7 \choose 2 $ $\times 2! $ = $30240$

share|improve this answer
    
hello sir harish i tried to compute 6! x 7P2 x 2! in my calculator but i got 60,480 ways? but when i divided it in 2 i got 30240 ways is this correct thanks :D –  Amelia Amore Sep 11 '13 at 9:22
    
no it is $7C2$ * 2! or just 7P2 . you have done both –  Harish Kayarohanam Sep 11 '13 at 9:23
    
so this mistake made ur answer double what I got –  Harish Kayarohanam Sep 11 '13 at 9:28
    
yeah i just used the 7P2 not the 7C2. thanks Sir, i got it now, hope this is correct. –  Amelia Amore Sep 11 '13 at 9:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.