Determining Combinations in Word Problems

[emilykschwartz]

I'm having difficulties with the problem below (and problems that are similar in setup). I stumbled across it in a practice GMAT from MBA.com:

A 4-person task force is to be formed from the 4 men and 3 women who work in Company G's human resources department. If there are to be 2 men and 2 women on this task force, how many different task forces can be formed?

Do I use a factorial to solve this problem?

[The correct answer is 18. I'm struggling to achieve it.] Help would be much appreciated.

[parthian7]

This is a classic combination problem..almost as basic as it gets..

Let's say
A: # of ways you can select 2 men out of a group of 4
B: # of ways you can select 2 women out of a group of 3
answer is A*B.

using combination it would (4 C 2)(3 C 2) = (4!/2!2!)(3) = 18

=============================================

If you don't wanna go by the formula, you can solve it using permutations:

Let's calculate for A first:
You have 2 spots to fill and you have 4 items to chose from.
Clearly, you will have 4 options for the 1st and 3 options for the 2nd spot. Therefore 4*3.
Mind you, you need to get rid of redundancies. 4*3 is the number of ways you can pick 2 men from a group of 4 and line them up if the order in which they stood in the line mattered. However, the order doesn't matter here. That is, it doesn't matter if you end up with John in the 1st spot and Dwayne in 2nd or vice versa. Therefore, you need to divide 4*3 by the number of ways 2 items in a line can be rearranged: 2!
so you'll have 4*3/2!=6.

similarly for B..
2 spots, 3 to choose from --> 3*2/2! --> 3
voila!

A*B = 18

hope it helps :)

[emilykschwartz]

This is very helpful. Forgive me for asking the following, which probably has a very obvious answer, but I'm pretty unfamiliar with this type of problem. We have --

"Let's say
A: # of ways you can select 2 men out of a group of 4
B: # of ways you can select 2 women out of a group of 3
answer is A*B.

using combination it would (4 C 2)(3 C 2) = (4!/2!2!)(3) = 18"

Could you walk me through (to the point of being painfully literal) how you solved this?

I get that (4 C 2) is supposed to represent the # of ways you can select 2 men out of a group of 4 and (3 C 2) is the # of ways you can select 2 women out of a group of 3. What I don't understand is how you went from that to (4!/2!2!)(3).

Thank you so, so much.

[parthian7]

sorry I assumed we know:

(n C n-1) = (n C 1) = n

(n C n-1) = n!/(n-1!)*1! = n
try doing the other one. I'm sure you can prove to yourself that it also yields n ;)

that's why I didn't expand (3 C 2). Just wrote 3 instead.

clear now ?

[RonPurewal]

emily, there are formulas that will do that.

with numbers this small, though, there's no reason to know formulas. if there are only a small number of possibilities, just MAKE A LIST!

if the 4 men are 'a', 'b', 'c', and 'd', then here are the possibilities for choosing two of them:
a, b
a, c
a, d
b, c
b, d
c, d
(six different ways)

if the 3 women are 'x', 'y', and 'z', then here are the possibilities for choosing two of them:
x, y
x, z
y, z
(three different ways)

so that's 6 x 3 = 18 ways in total.

[emilykschwartz]

Ron and Parthian --

Both of you have been very helpful! I dug up the chapter on Combinatorics and have delved into it a bit to gain a better understanding of the concept. Thanks so much for taking the time to respond! It's been much appreciated.

[RonPurewal]

glad to help.

[parthian7]

:)

[jnelson0612]

:-) Thanks, everyone!

[EstanisladoM601]

emily, there are formulas that will do that.

with numbers this small, though, there's no reason to know formulas. if there are only a small number of possibilities, just MAKE A LIST!

if the 4 men are 'a', 'b', 'c', and 'd', then here are the possibilities for choosing two of them:
a, b
a, c
a, d
b, c
b, d
c, d
(six different ways)

if the 3 women are 'x', 'y', and 'z', then here are the possibilities for choosing two of them:
x, y
x, z
y, z
(three different ways)

so that's 6 x 3 = 18 ways in total.

Ron,

Here is what I did:

4!/(2!(2!)) + 4!/(2!(1)) I got the same answer: 18. However, I noticed you multiplied when I added. Am I doing something wrong here?

Thanks in advance for your answer!

[RonPurewal]

4!/(2!(2!)) + 4!/(2!(1)) I got the same answer: 18. However, I noticed you multiplied when I added. Am I doing something wrong here?

Thanks in advance for your answer!

^^ the red "4" should be a "3" (you're choosing from only 3 women). ...and, yes, you need to multiply these numbers, since you're making BOTH of these sets of choices.

adding is what you do when there are two alternative sets of possibilities.
e.g., if there are 6 different ways to enter a house through the front door and 4 different ways to enter through the back door, then there are 10 (not 24) different ways to enter the house overall.

[RonPurewal]

what i'm wondering here is... is that how you ORIGINALLY worked the problem? i.e., without having seen an answer key, or any other indication that the answer should be 18?

...because it seems unlikely that you'd have made two different mistakes that, just coincidentally, "undo" each other and produce a nominally correct answer.

in other words, what i'm asking is -- when you came up with this, did you already know that the answer was supposed to be 18, and so were you just trying to combine steps (...in any way you could, basically) to give that answer?
or was this really just a giant coincidence?

[EstanisladoM601]

what i'm wondering here is... is that how you ORIGINALLY worked the problem? i.e., without having seen an answer key, or any other indication that the answer should be 18?

...because it seems unlikely that you'd have made two different mistakes that, just coincidentally, "undo" each other and produce a nominally correct answer.

in other words, what i'm asking is -- when you came up with this, did you already know that the answer was supposed to be 18, and so were you just trying to combine steps (...in any way you could, basically) to give that answer?
or was this really just a giant coincidence?

This was pure pressure, in the heat of the moment, man. A blessing in other words. I was taking a practice CAT [GMAT Prep], under official conditions. During the test, I did use the same setup that I showed you [with the two 4s], but initially, I multiplied the answers from each. That led me to a rather large answer, which wasn't one of the options. That's when I looked back at my work, and figured that I had made a mistake, thinking that I should've added instead of multiplied the answers from each section. It must have been a 300-500/500-600 level question, as they didn't work too hard on the trap answers.

It wasn't until I reviewed this problem, during my review of the CAT, that something inside me was questioning my methodology, even though I had gotten the correct answer. That's when I looked up this problem on the blog.

Thank you so much for your explanation. This does makes a lot more sense now. 1. I got the methodology correct for setting up each section [referring to the 4! vs 3!]. 2. I can see now that I need to understand the problem better to know when to add or multiply.

Stan

[RonPurewal]

you're welcome... but, for this problem (and others like it), it appears you're still missing what is THE MOST IMPORTANT POINT here BY FAR.

specifically:

I can see now that I need to understand the problem better to know when to add or multiply.

^^ for a problem like this one, you shouldn't be "adding" or "multiplying" at all. when it's obvious that there are only a small number of possibilities, you should FORGET ABOUT FANCY APPROACHES, and just LIST ALL THE POSSIBILITIES!

ironically, this is all you have to do to solve a solid majority of "harder" problems about combinations. generally speaking, the problems whose answers are too big for a simple list are actually the easier ones.

[EstanisladoM601]

Good to know. Thanks.