1. ## combinations

1)How many four letter words can be formed from the word proportion?

2)How many committees of 9 people can be chosen from 12, if two certain people must never be in the same committee?

1) p is repeated twice, as is r, and o is repeated three times

so would it be 10C4 / (2!2!3!) to get rid of the double ups?

2) 1,2,3,4,5,6,7,8,9,10,11,12

lets say 1 is taken out, u have 11C9, and the same if two was taken out 11C9, and for the case where they are both together:10C9??

I need some help with this one.

2. Originally Posted by Gul
1)How many four letter words can be formed from the word proportion?

2)How many committees of 9 people can be chosen from 12, if two certain people must never be in the same committee?
1)2 P's,3 O's,2 R's,1 T,1 I,1 N

Case 1 All letters are different=$\displaystyle \binom{6}{4}(4!)$

Case 2 Two letters same and two different=$\displaystyle \binom{3}{1}.\binom{5}{2}.\frac{4!}{2!}$

Case 3 Two letters same and two letters same=$\displaystyle \binom{3}{2}.\frac{4!}{2!2!}$

Case 4 Three letters same and one different==$\displaystyle \binom{5}{1}.\frac{4!}{3!}$

Total number of words=Add up all the cases

2)No. of ways to make the committee=$\displaystyle \binom{12}{9}$

No. of ways in which the committee can be formed so that the two persons are in the same committee=$\displaystyle \binom{10}{7}$

No. of ways in which the committee can be made so that the two persons are not in the same committee=$\displaystyle \binom{12}{9}-\binom{10}{7}$

3. Hello, Gul!

2) How many committees of 9 people can be chosen from 12 people,
if two certain people must never be in the same committee?
There is a back-door approach to this problem . . .

With no restrictions, there are: .$\displaystyle _{12}C_9 \:=\:220$ possible committees.

Let the two anti-social people be $\displaystyle A$ and $\displaystyle B.$

How many committees have both $\displaystyle A$ and $\displaystyle B$ together?
Place $\displaystyle A$ and $\displaystyle B$ on the committee.
Choose the other 7 members from the remaining 10 people: .$\displaystyle _{10}C_7 \:=\:120$ ways.
. . Hence, there are 120 committees with both $\displaystyle A$ and $\displaystyle B.$

Therefore, there are: .$\displaystyle 220 - 120 \:=\:\boxed{100}$ committees
. . in which $\displaystyle A$ and $\displaystyle B$ are not together.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This problem can be solved head-on . . . It just takes longer.

There are three cases to consider.

(1) $\displaystyle A$ is on the committee, hence $\displaystyle B$ is not.
. . .We choose the other 8 members from the remaining 10 people.
. . .There are: .$\displaystyle _{10}C_8 \:=\:45$ ways.

(2) $\displaystyle B$ is on the committee, hence $\displaystyle A$ is not.
. . .We choose the other 8 members from the remaining 10 people.
. . .There are: .$\displaystyle _{10}C_8 \:=\:45$ ways.

(3) Neither $\displaystyle A$ nor $\displaystyle B$ is on the committee.
. . .Choose the 9 members from the other 10 people.
. . .There are: .$\displaystyle _{10}C_9 \:=\:10$ ways.

Therefore, there are: .$\displaystyle 45 + 45 + 10 \:=\:\boxed{100}$ committees
. . in which $\displaystyle A$ and $\displaystyle B$ are not together.

But I don't understand why you have to multiply by 4!, 4!/2! etc?

I thought about them again, for the first one:

I have 6 letters available pick one: 6C1*1C1, then 5C1*1C1, then 4C1*1C1, then 3C1*1C1, so thats 6*5*4*3 = 360
that's the same as 6C4*4! = 360

For the second situation:two same, two different

(3C1*2C2*5C1*1C1*4C1*1C1)/2! ?

^^3 ways of picking pairs, pick both cards from that pair, 5 types of letters available pick one, then four pick one?

5. my dear,you are also required to arrange the letters to form the words.Simply selection will not do