Permutations and Combinations: Question 2

Worked solution

Part (a): No restriction

There are $6$ distinct friends to arrange in $6$ positions in a row, so the number of arrangements is

$6! = 6\times5\times4\times3\times2\times1 = 720.$

Part (b): Reyna and Soren together (block method)

Since Reyna and Soren must be next to each other, glue them together into a single “block”. This leaves $5$ objects to arrange in a row: the [Reyna–Soren] block together with the other $4$ friends (Priya, Quincy, Tavita, Uche).

$5! = 120 \text{ ways to arrange the } 5 \text{ objects.}$

Within the block, Reyna and Soren can swap places ($2!=2$ ways: Reyna-then-Soren, or Soren-then-Reyna):

$5!\times2! = 120\times2 = 240.$

Check (direct position-counting): in a row of $6$ seats, the adjacent seat-pairs are $(1,2),(2,3),(3,4),(4,5),(5,6)$, that’s $5$ adjacent pairs. For each pair, Reyna and Soren can sit in $2$ orders, and the other $4$ friends fill the remaining $4$ seats in $4!=24$ ways: $5\times2\times24 = 240.$ Both methods agree: $240$.

Part (c): Reyna and Soren not together (complement)

The arrangements where Reyna and Soren are not next to each other are exactly the total arrangements minus the ones where they are next to each other (from part (b)):

$720 - 240 = 480.$

Check (direct counting): the number of ways to place two distinct people (Reyna, Soren, in order) into $6$ seats is $^6P_2 = 6\times5=30$. Of these, the adjacent placements found in part (b)‘s check are $5\times2=10$. So non-adjacent placements of the pair number $30-10=20$. The remaining $4$ friends fill the other $4$ seats in $4!=24$ ways: $20\times24 = 480.$ Both methods agree: $480$.

Part (d): Tavita at one of the two ends

There are $2$ choices for which end Tavita occupies (the front of the row or the back of the row). Once Tavita’s seat is fixed, the other $5$ friends fill the remaining $5$ seats in $5!$ ways:

$2\times5! = 2\times120 = 240.$

Check (split into the two cases): if Tavita is at the front seat, the other $5$ friends fill the remaining $5$ seats in $5!=120$ ways; if Tavita is at the back seat, likewise $5!=120$ ways. These two cases can’t both happen at once, so they simply add: $120+120 = 240.$ Both methods agree: $240$.

Final answers

• (a) Total arrangements $= \boxed{720}$
• (b) Reyna and Soren together $= \boxed{240}$
• (c) Reyna and Soren not together $= \boxed{480}$
• (d) Tavita at one of the two ends $= \boxed{240}$