Permtations and Combinations Word problems

[Louise Johnson]

Question #1
Four men and five women form the board of directors of a bank. In how many ways can a five-member subcommitte be formed from this board if:
a) the women must have a majoity on this subcommitee.

my answer:
case 1 five women
\(\displaystyle \L\\\frac{{5!}}{{0!5!}} =\: _5 C_5 = 1\)

case 2 four women and one man
\(\displaystyle \L\\\frac{{5!}}{{1!4!}} \times \frac{{4!}}{{5!1!}} = \:_5 C_4 \times _4 C_1 = 20\)

Add the totals together and get a final answer of 21

b) There must be at least one man on this subcommittee

my answer:
All possible - no men on the subcommitee

\(\displaystyle \L\\_9 C_5 - _5 C_5 = 125\)

I can only hope I am doing these correctly
Thank you
Sincerly
Louise

 

[galactus]

Question #1
Four men and five women form the board of directors of a bank. In how many ways can a five-member subcommitte be formed from this board if:
a) the women must have a majoity on this subcommitee.

my answer:
case 1 five women
\(\displaystyle \L\\\frac{{5!}}{{0!5!}} =\: _5 C_5 = 1\)

case 2 four women and one man
\(\displaystyle \L\\\frac{{5!}}{{1!4!}} \times \frac{{4!}}{{5!1!}} = \:_5 C_4 \times _4 C_1 = 20\)

Add the totals together and get a final answer of 21

Looks good, except you forgot the case when there are 3 women and 2 men.

b) There must be at least one man on this subcommittee

my answer:
All possible - no men on the subcommitee

\(\displaystyle \L\\_9 C_5 - _5 C_5 = 125\)

I can only hope I am doing these correctly
Thank you
Sincerly
Louise

Yep, took the case where there are no men and subtracted from the total number of ways. Good.

 

[soroban]

[soze=14]Hello, Louise![/size]

Four men and five women form the board of directors of a bank.
In how many ways can a five-member subcommitte be formed from this board if:

a) the women must have a majoity on this subcommitee.

my answer:

case 1: five women
\(\displaystyle \L\\\frac{{5!}}{{0!5!}} =\: _5 C_5 = 1\)

case 2: four women and one man
\(\displaystyle \L\\\frac{{5!}}{{1!4!}} \times \frac{{4!}}{{3!1!}} = \:_5 C_4 \times _4 C_1 = 20\)

Add the totals together and get a final answer of 21

You forgot . . .

case 3: three women and two men

\(\displaystyle \L\frac{5!}{2!3!}\,\times\,\frac{4}{2!2!} :=\:10\,\times\,6\:=\:60\) more cases.

b) There must be at least one man on this subcommittee

my answer:

All possible - no men on the subcommitee

\(\displaystyle \L\\_9 C_5 - _5 C_5 = 125\;\;\) . . . Right!

 

[Louise Johnson]

Thanks for helping me out. It feels good knowing I was pretty darn close!
Thanks again
Louise