# In how many ways we can arrange 12 people in a row if 5 men are constrained to sit next to each other together?

In how many ways we can arrange $12$ people in a row if $5$ are men and they must sit next to each other?

My approach

I consider $5$ men as one entity and so now there are $8$ people to be seated in a row, which is done in 8! ways. The $5$ men considered as one entity can themselves be seated in $5!$ ways. So the total number of ways are $8!5!$ (multiplication rule). What's wrong with the approach?

• Who said your approach is wrong ? – Aakash Kumar Jul 25 '16 at 15:44
• Well it's obviously wrong because $12 - 5 \neq 8$... – Ashvin Swaminathan Jul 25 '16 at 15:45
• which amounts to 8!5! – Satish Ramanathan Jul 25 '16 at 15:49
• @AshvinSwaminathan, the grouped entity is the eighth figuratively – Satish Ramanathan Jul 25 '16 at 15:51
• Your answer is fine. Another way you could do this is to first arrange the 7 other people, then choose the gap for the 5 men, and then arrange the 5 men in this gap, giving $7!\cdot8\cdot5!$ (which is the same as your answer). – user84413 Jul 25 '16 at 22:00

It makes sense to me. If there were only the 5 men present, there would be $5!$ choices. If there were 6 people present, the sixth person could sit either on the left side or the right, so $2! \times 5!$. And so on for $n$ total people, with $(n-4)!\times 5! .$ When $n=12$, we have $8!\times 5!$ different ways, as you say.