Math Help - Combination/ Permutation Question

1. Combination/ Permutation Question

I have a few questions i don't understand T_T

1. A committe of 4 people is to be chosen form a group of 8 people- 4 women and 4 men. How many ways can the committee be chosen so as to include exactly 3 women.

The answer is 70 but i keep getting 96. I don't understand how they got 70 =3=

2. From a deck of 52 cards, the 12 face cards and 4 aces are removed. From the 16 cards, 4 are chosen. How many different ways are there to choose the numbers?

3. In the Lotto 535 lottery, you must choose 4 numbers from 1 to 35 and one number from 1 to 25. How many different ways are there to choose the numbers?

4. Solve for n
nP2= 12
The answer is 4 but i get 5 =3=

n!/ n-2! = 12
then i expanded and canceled out that n! leaving me (n-2)(n-1) = 12
i expanded it then i get n^2- 3n+2 = 12
i got n= 5 and -2
D:
i don't know what i did wrong.

help anyone? (:

2. For (1) I get 16 since ${4\choose 3}{4\choose 1}=(4)(4)=16$

(4) IT's n(n-1)=12 producing $n^2-n-12=(n-4)(n+3)=0$ so n=4

3. Originally Posted by matheagle
I get 16
Ohh. o_o
weird T_T
i don't like these questions

4. so how did you get 96 and who says its 70?

5. Originally Posted by matheagle
so how did you get 96 and who says its 70?
O; oh em gee. nvm it's 16 D:
i was looking at another question =3=

buut. i think using the 8 instead the 4 thats why i was getting such big numbers.
but how do you know it's a combination or a permutation from

( 4 3 ) (4 1)

6. I thought it was 16 too.

${8\choose 4}=70$, which would be the number of ways to select 4 people out of 8 disregarding gender. Maybe you're reading the wrong answer?

Haha.. you just beat me to it.

It's a combination, because the order doesn't matter.

7. you want to pick 3 women from a group of 4.

that's ${4\choose 3}=4$

you can write out ABCD women, and see there are 4 ways.
BUT the easier way is to cast off one and see there are 4 cast offs, leaving you with 3 women
ABC sending D home alone
ABD sending off C
ACD sending off B
BCD sending off A

same with 1 man from 4, there are 4 men to choose from.

8. Originally Posted by matheagle
you want to pick 3 women from a group of 4.

that's ${4\choose 3}=4$

you can write out ABCD women, and see there are 4 ways.
BUT the easier way is to cast off one and see there are 4 cast offs, leaving you with 3 women
ABC sending D home alone
ABD sending off C
ACD sending off B
BCD sending off A

same with 1 man from 4, there are 4 men to choose from.
okkay. i think i got it xP

9. 2. It looks like their just asking for the number of ways to choose 4 cards out of the remaining 16, although I'm not too sure what is meant by "numbers."

Anyway, ${16 \choose 4} =...$

3. It's the number of ways to choose 4 numbers out of 35 multiplied by the number of ways to choose 1 number out of 25

10. Originally Posted by downthesun01
2. It looks like their just asking for the number of ways to choose 4 cards out of the remaining 16, although I'm not too sure what is meant by "numbers."

Anyway, ${16 \choose 4} =...$
okay...