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How many ways can you arrange 8 people at a round table (clockwise and anticlockwise arrangements are considered different)?

Mathematics

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Answers

3 parts
a
330
b
150
c
100

Step-by-step solution

9 steps

How many ways can you form a 4-person committee from 6 men and 5 women?
Total number of people = 6 men + 5 women = 11 people.
The committee size is 4 people.

Part a· 3 steps

With no restrictions

  1. Determine the total number of people (nn) and the committee size (kk).

    n=11n = 11 (total people) k=4k = 4 (committee members)

  2. Use the combination formula, as the order of selection does not matter for a committee.

    C(n,k)=(nk)=n!k!(n−k)!C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}

  3. Substitute the values and calculate.

    C(11,4)=11!4!(11−4)!=11!4!7!C(11, 4) = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} C(11,4)=11×10×9×8×7!4×3×2×1×7!C(11, 4) = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} C(11,4)=11×10×9×84×3×2×1C(11, 4) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} C(11,4)=792024C(11, 4) = \frac{7920}{24} C(11,4)=330C(11, 4) = 330

    The number of ways to form a 4-person committee with no restrictions is 330\boxed{330}.

Part b150

the committee has 2 men and 2 women

  1. Determine the number of ways to choose 2 men from 6 men.

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

  2. Determine the number of ways to choose 2 women from 5 women.

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

  3. Multiply the number of ways to choose men and women, as these are independent selections.

    Total ways = C(6,2)×C(5,2)=15×10=150C(6, 2) \times C(5, 2) = 15 \times 10 = 150

    The number of ways to form a committee with 2 men and 2 women is 150\boxed{150}.

Part c100

the committee has only 1 woman

  1. If the committee has only 1 woman, and the committee size is 4, then the remaining 3 members must be men.

    Determine the number of ways to choose 1 woman from 5 women.

    C(5,1)=5!1!(5−1)!=5!1!4!C(5, 1) = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} C(5,1)=5×4!1×4!=5C(5, 1) = \frac{5 \times 4!}{1 \times 4!} = 5

  2. Determine the number of ways to choose 3 men from 6 men.

    C(6,3)=6!3!(6−3)!=6!3!3!C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} C(6,3)=6×5×4×3!3×2×1×3!=1206=20C(6, 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 and men.

    Total ways = C(5,1)×C(6,3)=5×20=100C(5, 1) \times C(6, 3) = 5 \times 20 = 100

    The number of ways to form a committee with only 1 woman is 100\boxed{100}.

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