Can somebody help me with this problem , please?


Of the twelve participants in a certain competition, half are male, and half of the males are younger than 18 years of age. If half of the female competitors are also younger than 18 years of age, into how many distinct groups of 4 competitors could the participants be divided if each group must contain two males under 18 years of age and 2 females over 18 years of age?
A. 3
B. 4
C. 6
D. 9
E. 20
Answer:9

Are you asking how to do this? It's rather straightforward so maybe you're just posting it for others?

Are you asking how to do this? It's rather straightforward so maybe you're just posting it for others?

Yes, please help me. I don't find it straight forward at all!:confused:

Oh, okay. My advice to most people on this forum having trouble with stats problems is to just forget about permutation and combination formulas. They are probably going to do more confusing than anything else and can be a bit advanced for the purposes of the DAT and the backgrounds of most pre-dental students. Just think through these logically. So you have 3 males under 18 and 3 females over 18. Think of the males as A, B, and C ... and the females as 1, 2, and 3.

Here are the possibilities.

AB1
BC1
AC1

and the same for females 2 and 3:

AB2
BC2
AC2

AB3
BC3
AC3

3 sets of 3 for a total of 9.

make sense?

Answer is D, yes?

Oh, okay. My advice to most people on this forum having trouble with stats problems is to just forget about permutation and combination formulas. They are probably going to do more confusing than anything else and can be a bit advanced for the purposes of the DAT and the backgrounds of most pre-dental students. Just think through these logically. So you have 3 males under 18 and 3 females over 18. Think of the males as A, B, and C ... and the females as 1, 2, and 3.

Here are the possibilities.

AB1
BC1
AC1

and the same for females 2 and 3

3 sets of 3 for a total of 9.

make sense?


Wow, that's way more understandable than my Kaplan explanation! Thanks!

Well here's the combo/perm answer in case you want it.

There are 4 equal groups of 3 if you divide them up into male/female and over/under 18.

You want 2 males from one of those groups and 2 females from another group. So you choose your two males using (3 choose 2) and your two females using (3 choose 2). Multiply them together and get 9.


===

Thebozz you need to have 2 females per group, not 1. Same answer though.

Thanks Streetwolf, you're right.

So the possibilities are:

A,B,1,2
B,C,1,2
A,C,1,2

and the same for females 2,3 and 1,3:

A,B,2,3
B,C,2,3
A,C,2,3

A,B,1,3
B,C,1,3
A,C,1,3

Thanks Streetwolf, you're right.

So the possibilities are:

A,B,1,2
B,C,1,2
A,C,1,2

and the same for females 2,3 and 1,3:

A,B,2,3
B,C,2,3
A,C,2,3

A,B,1,3
B,C,1,3
A,C,1,3

How would you use the formulas if you wanted to?

How would you use the formulas if you wanted to?

Read my post above.