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How many ways can you arrange 8 people at a round table (clockwise and anticlockwise arrangements are considered different) with no restrictions?

Mathematics

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Step-by-step solution

8 steps

1. How many ways can you form a 4-person committee from 6 men and 5 women?

Part a· 2 steps

with no restrictions

We need to choose 4 people from a total of 6+5=116 + 5 = 11 people. Since the order does not matter, we use combinations.

  1. Calculate the total number of people.

    Total people = 6 men+5 women=11 people6 \text{ men} + 5 \text{ women} = 11 \text{ people}

  2. Use the combination formula (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!} to choose 4 people from 11.

    (114)=11!4!(11−4)!=11!4!7!\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} =11×10×9×8×7!4×3×2×1×7!= \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} =11×10×9×84×3×2×1= \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} =11×10×3=330= 11 \times 10 \times 3 = 330

    The number of ways is 330.

Part b· 3 steps

the committee has 2 men and 2 women

We need to choose 2 men from 6 men AND 2 women from 5 women.

  1. Choose 2 men from 6 men.

    (62)=6!2!(6−2)!=6!2!4!=6×5×4!2×1×4!=302=15\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15

  2. Choose 2 women from 5 women.

    (52)=5!2!(5−2)!=5!2!3!=5×4×3!2×1×3!=202=10\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10

  3. Multiply the number of ways to choose men by the number of ways to choose women.

    15×10=15015 \times 10 = 150

    The number of ways is 150.

Part c· 3 steps

the committee has only 1 woman

If the committee has 1 woman, and it's a 4-person committee, then the remaining 4−1=34 - 1 = 3 members must be men.

  1. Choose 1 woman from 5 women.

    (51)=5!1!(5−1)!=5!1!4!=5×4!1×4!=5\binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = \frac{5 \times 4!}{1 \times 4!} = 5

  2. Choose 3 men from 6 men.

    (63)=6!3!(6−3)!=6!3!3!=6×5×4×3!3×2×1×3!=1206=20\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = \frac{120}{6} = 20

  3. Multiply the number of ways to choose women by the number of ways to choose men.

    5×20=1005 \times 20 = 100

    The number of ways is 100.

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