Probability/Combinations - Question

• Nov 27th 2008, 08:04 AM
cnmath16
Probability/Combinations - Question
A club consists of 25 married couples. Determine how many ways a committee of 7 people can be formed from these couples if..

c) at least one person must be a man

d) there must be at least 3 women and at least 2 men

my answer: (25 3) (25 7-2)
= 25 Cr 3 x 25 Cr 7-2
= 1,105,598,500

Is this correct?
• Nov 27th 2008, 09:13 AM
Soroban
Hello, cnmath16!

Quote:

A club consists of 25 married couples.
Determine how many ways a committee of 7 people can be formed from these couples if:

c) at least one person must be a man.

There are: . ${50\choose 7} \:=\:99,\!884,\!400$ possible committees.

The opposite of "at least one man" is "no men" (all women).

There are: . ${25\choose7} \:=\:480,\!700$ all-women committees.
. . Hence, the rest must contain at least one man.

Answer: . $99,\!884,\!400 - 480,\!700 \;=\;99,\!403,\!700$ committees with at least one man.

Quote:

d) there must be at least 3 women and at least 2 men.
There are three cases: . $\begin{array}{ccc}(1) & \text{3 women, 4 men} \\ (2) & \text{4 women, 3 men} \\ (3) & \text{5 women, 2 men} \end{array}$

. . $\begin{array}{ccccc}\text{(1) 3 women, 4 men:} & {25\choose3}{25\choose4} &=& 29,\!095,\!000 \\ \\[-3mm]
\text{(2) 4 women, 3 men:} & {25\choose4}{25\choose3} &=& 29,\!095,\!000 \\ \\[-3mm]
\text{(3) 5 women, 2 men:} & {25\choose5}{25\choose2} &=& 15,\!939,\!000 \end{array}$

Answer: . $29,\!095,\!000 + 29,\!095,\!000 + 15,\!939,\!000 \;=\; 74,\!129,\!000$ committees.