# Thread: Permutation word problem

1. ## Permutation word problem

Ms. Brown has just learned that she can send as many participants as she likes to the Explorer Post roundtable, but she must send at least one girl. Ms. Brown has six boys and four girls in her post. How many choices does she have in this circumstance?

2. Originally Posted by ihavvaquestion
Ms. Brown has just learned that she can send as many participants as she likes to the Explorer Post roundtable, but she must send at least one girl. Ms. Brown has six boys and four girls in her post. How many choices does she have in this circumstance?
Note: This problem is not about permutations as you titled it.

Whenever people are used in combinations problems, they are usually assumed to be non-identical, for obvious reasons. (People are unique.)

With that assumption, the simplest approach I can see is to consider all possible ways to choose just the girl(s), and multiply by the number of ways to choose the boys.

A straightforward way to count ways to choose the girls is just C(4,1) + C(4,2) + C(4,3) + C(4,4).

This is the same as 2^4 - C(4,0) = (2^4)-1. Can you see why?

Then we multiply this by the number of ways to choose boys, which is 2^6. Can you see why?

So the overall answer is ((2^4)-1)*(2^6).

3. Sorry I got the title wrong...

I don't think what you are doing is right though. I am sure the boys and girls are identified FOR obvious reasons.

So the total number of combinations would be (2^10) - 1; because if one girl has to be sent, then she cannot send zero people...hence the minus 1. Then I think I would subtract all combinations that included only boys...correct???

So I am not sure if this is correct, but would 2^10 - 2^6 -1, or 959 combinations be correct?

4. Originally Posted by ihavvaquestion
Sorry I got the title wrong...

I don't think what you are doing is right though. I am sure the boys and girls are identified FOR obvious reasons.

So the total number of combinations would be (2^10) - 1; because if one girl has to be sent, then she cannot send zero people...hence the minus 1. Then I think I would subtract all combinations that included only boys...correct???

So I am not sure if this is correct, but would 2^10 - 2^6 -1, or 959 combinations be correct?
I'm so mad at Firefox because I typed a whole response and it erased it!

Your method is fine but it subtracts out the case of sending no students twice. You should instead write 2^10 - (2^6 - 1) - 1 = 960, which is the same as ((2^4)-1)*(2^6) = 15*64 = 960.

And there seems to have been a slight communication error regarding the word "identical" which you wrote as "identified" for some reason. If boys and girls were considered identical for this problem, the answer would be 4*7 = 28, because you could choose between 1 and 4 girls inclusive, and between 0 and 6 boys inclusive, and it doesn't matter which girls or boys because they are identical. But, as I said, we assume that the people are non-identical, as is customarily done.

5. ok thanks for your help on this, but i am still a little confused.

I guess I am not understanding if the answer is 960, 959

Shouldnt it be the total combinations possible minus the total combinations (that include boys only) minus 1; there is only 1 way to send zero people right?

(2^10) - (2^6) -1 =959

6. Originally Posted by ihavvaquestion
ok thanks for your help on this, but i am still a little confused.

I guess I am not understanding if the answer is 960, 959

Shouldnt it be the total combinations possible minus the total combinations (that include boys only) minus 1; there is only 1 way to send zero people right?

(2^10) - (2^6) -1 =959
2^6 includes the combination consisting of sending no students. When you take (2^10) - (2^6) - 1, you subtract this combination twice.

Consider a smaller example.

2 girls named Abigail and Betty, 3 boys named Charlie, Dan, and Edward. There are 2^3 combinations in which all students are boys, but this includes the empty set which is vacuously true.

{empty set}
C
D
E
C, D
C, E
D, E
C, D, E

As you can see, if you exclude the empty set then there are 2^3 - 1 ways to choose only boys, where at least one student is chosen.

So the correct answer is 960, not 959.

7. okay, so could I just say that there are 2^10 - 2^6 combinations, since the empty set is being subtracted out in the 2^6?

8. Originally Posted by ihavvaquestion
okay, so could I just say that there are 2^10 - 2^6 combinations, since the empty set is being subtracted out in the 2^6?
Right.