combination problem

[flyingfreedom]

A coach needs to choose an 8-member volleyball team from 10 males and 12 females.
If there must be at least 3 of each gender on the team, how many different teams
are possible?

A. 103 950
B. 150 480
C. 254 430
D. 319 770


I tried solving with combinations I did 10C3*12C3*16C2
unfortunately that points to a number way too large.
So I know you use combinations, but obviously the way I'm doing it isn't working.
The answer is apparently C, 254 430.

Thanks in advance.

 

[soroban]

Hello, flyingfreedom!

galactus is absolutely correct . . . I'll walk you through it.

A coach needs to choose an 8-member volleyball team from 10 males and 12 females.
If there must be at least 3 of each gender on the team, how many different teams are possible?

. . $\displaystyle (A)\;103,\!950 \qquad (B)\;150,\!480 \qquad (C)\;254,\!430 \qquad (D)\;319,\!770$

There is no one formula for the answer; we must consider the three possible cases.


3 males, 5 females

. . \(\displaystyle \text{There are: }\:\begin{array}{ccc}{10\choose3}&=& 120\text{ ways to choose 3 males} \\ \\[-3mm] {12\choose5} &=& 792\text{ ways to choose 5 females}\end{array}\)

. . $\displaystyle \text{Hence, there are: }\;120 \times 792 \:=\:95,\!040\text{ ways to choose 3 males and 5 females.}$


4 males, 4 females

. . \(\displaystyle \text{There are: }\:\begin{array}{ccc}{10\choose4} &=& 210\text{ ways to choose 4 males} \\ \\[-3mm]{12\choose4} &=& 495\text{ ways to choose 4 females}\end{array}\)

. . $\displaystyle \text{Hence, there are: }\:210 \times 495 \:=\:103,\!950\text{ ways to choose 4 males and 4 females.}$


5 males, 3 females

. . \(\displaystyle \text{There are: }\:\begin{array}{ccc}{10\choose5} \:=\:252\text{ ways to choose 5 males} \\ \\[-3mm] {12\choose3} \:=\:220 \text{ ways to choose 3 females} \end{array}\)

. . $\displaystyle \text{Hence, there are: }\:252 \times 220 \:=\: 55,440\text{ ways to choose 5 males and 3 females.}$



\(\displaystyle \text{Therefore, there are: }\:95,\!040 + 104,\!950 + 55,\!440 \;=\;\boxed{254,\!430\text{ ways}} \quad\hdots\quad \text{answer (C)}\)

 

[flyingfreedom]

Thanks a lot there Soroban. I learn a lot from your walk throughs!

 

[galactus]

Last time I answer anything for you