# Perms & Coms (1 Viewer)

#### Jacobagel

##### New Member
Unable to do the following question. Could someone please provide an explanation:
Eight people are to form two queues of four. In how many ways can this be done if Sean and Liam must stand in the same queue?

#### fan96

##### 617 pages
If the queues are distinct, then we should have

$\bg_white 17280 = \underbrace{\left( \binom{8}{4} \times 4! \times 4!\right)}_{\text{no restrictions}} - \underbrace{\left(\binom{6}{3} \times 4!\ \times 4!\right)\times 2}_{\text{Sean and Liam in diff. queues}},$
or

$\bg_white 17280 = \left( \binom{6}{4} \times 4! \times 4!\right) \times 2.$

(Fix Sean and Liam in one queue, then pick four people from the remaining six for the other queue. Arrange $\bg_white 4!$ ways for each queue, and finally double to account for the distinct queues.)