Circular Permutations: 12 Practice Problems with Solutions

Fix the first person’s seat. The other 5 people fill the remaining seats in a definite order, starting from that person and going clockwise:

$$(6-1)!=5!=120.$$

Check: there are $6!=720$ ways in a row, and each circular seating appears 6 times among them (once for each rotation), so $720/6=120$.

Answer: 120

No. With numbered chairs, moving everyone changes who sits in chair 1, so rotations are different seatings. The round shape no longer matters:

$$6!=720.$$

Answer: 720

Around a circle there are $(7-1)!=720$ arrangements. Flipping the necklace turns each arrangement into its mirror image, and the two are now the same necklace (for 3 or more beads they are always different arrangements), so we divide by 2:

$$\frac{6!}{2}=360.$$

Answer: 360

$$\frac{(5-1)!}{2}=\frac{24}{2}=12.$$

Answer: 12

Treat Ann and Ben as one block. Now 7 units sit around the table: $(7-1)!=720$ ways. Inside the block they can sit as Ann–Ben or Ben–Ann: 2 ways.

$$720\times 2=1440.$$

Answer: 1440

All seatings: $(8-1)!=5040$. Seatings with Ann and Ben together: 1440 (Problem 5). So:

$$5040-1440=3600.$$

Answer: 3600

Seat the 5 men around the table: $(5-1)!=24$ ways. This leaves exactly 5 gaps between them, and these gaps are now distinguishable (each lies between two specific men). The 5 women fill them in $5!=120$ ways.

$$24\times 120=2880.$$

A common mistake is $(5-1)!\times(5-1)!$: the circle’s symmetry is used up once, by the men.

Answer: 2880

The 4 couples, as blocks, go around the table in $(4-1)!=6$ ways. Each couple can sit in 2 orders, independently: $2^4=16$.

$$6\times 16=96.$$

Answer: 96

Choose the 6 people: $\binom{10}{6}=210$. Seat them around the table: $(6-1)!=120$.

$$210\times 120=25{,}200.$$

Equivalently, there are $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5=151{,}200$ ordered rows of 6, and each circular seating corresponds to 6 rows: $151{,}200/6=25{,}200$.

Answer: 25,200

Seat the 5 other people around the table: $(5-1)!=24$ ways. They create 5 gaps. A, B and C must go into 3 different gaps (two in the same gap would be neighbors), and the order matters: $5\cdot 4\cdot 3=60$ ways.

$$24\times 60=1440.$$

Answer: 1440

Fixing the host’s seat removes the rotations. The guest of honor sits to the host’s left or right: 2 choices. The remaining 8 people fill the 8 remaining seats: $8!=40{,}320$.

$$2\times 40{,}320=80{,}640.$$

This matches the block method: $(9-1)!\times 2=80{,}640$.

Answer: 80,640

Fix Ann’s seat and number the other seats 1 to 6 going clockwise. Ben in seat 3 leaves seats 1 and 2 between them on one side. Ben in seat 4 leaves seats 5 and 6 between them on the other side. Any other seat gives a different number. So Ben has 2 choices, and the other 5 people fill the rest in $5!=120$ ways.

$$2\times 120=240.$$

Answer: 240