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Jessiereerosel19go
Answered

1. In how many ways can a committee of 6 persons be chosen from 7 single ladies and 8 gentlemen if there is no restriction on the gender?


2. In how many ways can a committee of 6 persons be chosen from 7 single ladies and 8 gentlemen if there must be 4 single ladies and 2 gentlemen?


3. In how many ways can a committee of 6 persons be chosen from 7 single ladies and 8 gentlemen if all are single ladies?

Sagot :

KAntoineDoixgo

✒️COMBINATIONS

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad \Large \:\: \rm 1) \; 5005 \: ways [/tex]

[tex] \qquad \Large \:\: \rm 2) \; 980 \: ways [/tex]

[tex] \qquad \Large \:\: \rm 3) \; 7 \: ways [/tex]

*Please read and understand my solution. Don't just rely on my direct answer*

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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]

Number 1:

No restrictions of gender means all can be participated. In a total of 15 people, get the combinations if it is picked 6 at a time.

[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm _nC_r = \frac{n!}{r!(n-r)!}} \end{align} [/tex]

  • [tex] \rm _{15}C_6 = \frac{15!}{6!(15-6)!} \\ [/tex]

  • [tex] \rm _{15}C_6 = \frac{15!}{6!\,9!} \\ [/tex]

  • [tex] \rm _{15}C_6 = \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot \cancel{9!}}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot \cancel{9!}} \\ [/tex]

  • [tex] \rm _{15}C_6 = \frac{\cancel{15} \cdot \cancel{14}^{\;7} \cdot 13 \cdot \cancel{12} \cdot 11 \cdot \cancel{10}^{\;5}}{\cancel{\,6\,} \cdot \cancel{\,5\,} \cdot \cancel{\,4\,} \cdot \cancel{\,3\,} \cdot \cancel{\,2\,}} \\ [/tex]

  • [tex] \rm _{15}C_6 = 7 \cdot 13 \cdot 11 \cdot 5 [/tex]

  • [tex] \rm _{15}C_6 = 5005 [/tex]

Therefore, there are 5005 ways to choose a committee of 6 person.

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Number 2:

First, determine the number of combinations of 7 ladies picked 4 at a time to be the members of the committee.

  • [tex] \rm _7C_4 = \frac{7!}{4!(7-4)!} \\ [/tex]

  • [tex] \rm _7C_4 = \frac{7!}{4! \, 3!} \\ [/tex]

  • [tex] \rm _7C_4 = \frac{7 \cdot 6 \cdot 5 \cdot \cancel{4!}}{\cancel{4!} \cdot 3 \cdot 2} \\ [/tex]

  • [tex] \rm _7C_4 = \frac{7 \cdot \cancel6 \cdot 5}{\cancel{3} \cdot \cancel2} \\ [/tex]

  • [tex] \rm _7C_4 = 7 \cdot 5 [/tex]

  • [tex] \rm _7C_4 = 35 [/tex]

There are 35 ways in which 4 ladies can be part of the committee. Now determine the number of combinations of 8 gentlemen picked 2 at a time to be the members of the committee.

  • [tex] \rm _8C_2 = \frac{8!}{2!(8-2)!} \\ [/tex]

  • [tex] \rm _8C_2 = \frac{8!}{2!\,6!} \\ [/tex]

  • [tex] \rm _8C_2 = \frac{8 \cdot 7 \cdot \cancel{6!}}{2 \cdot \cancel{6!}} \\ [/tex]

  • [tex] \rm _8C_2 = \frac{\cancel8^{\,4} \cdot 7}{\cancel2} \\ [/tex]

  • [tex] \rm _8C_2 = 4 \cdot 7\\ [/tex]

  • [tex] \rm _8C_2 = 28 [/tex]

There are 28 ways in which 2 gentlemen can be part of the committee. By applying FCP, determine the number of combinations if 4 ladies and 2 gentlemen can be part of the committee.

  • [tex] \implies \rm _7C_4 \cdot _8C_2 [/tex]

  • [tex] \implies \rm 35 \cdot 28 [/tex]

  • [tex] \implies \rm 980 [/tex]

Therefore, there are 980 ways in which 4 ladies and 2 gentlemen are the members of the committee.

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Number 3:

Determine the number of combinations of 7 ladies picked 6 at a time to be the members in a committee.

  • [tex] \rm _7C_6 = \frac{7!}{6! (7-6)!} \\ [/tex]

  • [tex] \rm _7C_6 = \frac{7!}{6!\,1!} \\ [/tex]

  • [tex] \rm _7C_6 = \frac{\,7 \cdot \cancel{6!}}{\cancel{6!}} \\ [/tex]

  • [tex] \rm _7C_6 = 7 [/tex]

Therefore, there are 7 ways in which all ladies are the members of the committee

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(ノ^_^)ノ [tex] \large\qquad\qquad\qquad\tt 3/1 /202 2[/tex]

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