Could somebody help me with this finite math problem?

Factual Questions
1

OK - I have worked two thirds of this problem and I am fairly sure I am correct so far, but for some reason I can’t get the third part to work. Here’s the question -
A group consists of 5 men and 8 women. A committee of 4 is to be formed from this group, and policy dictates that at least 1 woman be on this committee.

a) how many committees can be formed that contain exactly 1 man?
Answer I got - 280
b) How many committees can be formed that contain exactly 2 women?
Answer I got - 280
c) How many committees can be formed that contain at least 1 man?
This is where I get stuck.

I got the first two answers by using the combination formula:
for answer c:
c(5,1) * c(8,3) = 556 = 280
for answer b:
c(5,2) * c(8,2) = 1028 = 280
wouldn’t answer c be c(5,3) * c(8,1) + a + b?
I know the answer for c is 640.

so finally can somebody at least give me a hint to at least go about this problem, am I even working the first two parts correctly? Thanks for your help

2

I think it is:

([sub]5[/sub]C[sub]1[/sub][sub]8[/sub]C[sub]3[/sub]) + ([sub]5[/sub]C[sub]2[/sub][sub]8[/sub]C[sub]2[/sub]) + ([sub]5[/sub]C[sub]3[/sub]*[sub]8[/sub]C[sub]1[/sub]).

Alternatively, since you’re looking for committes that contain at least one man, the only alternative is that the committee contain all women, so you could figure out that number and subtract it from the total of all possible committees (containing at least one woman). Gives you the same answer.

3

Yet another way to go about it: A committee with at least one man must have exactly one man, exactly two men, or exactly three men (four men is forbidden by policy). You already know how many comittees there are with exactly one man (part a), and how many with exactly two men (same as exactly two women, part b). So if you find how many comittees there are with exactly three men, and add those three numbers together, you’ll have the number that can have at least one man.

4

Pardon my simplistic notation and explanations.

a) A committee with exactly one man must contain 3 women out of 8, and the man can be any one of the 5: (8!/(8-3)!) * 5 = 1680 unique committees

b) In a committee with exactly two women: the first woman can be one of eight, and the second is one fo the remaining seven. The two men, likewise, will be one of 5 and one of the remaining four. 87 * 54 = 1120

c) For the number of committees that contain men, subtract the number of all female committees [8!/(8-4!) = 1680] from the total number of possible committees (13! / (13-4)!) =17160) for a total of 15480

I’m assuming this isn’t homework…

5

Ignore the above. I somehow skipped the sentence on policy. My bad. My terrible, in fact.

6

For question C

Number of committees with only men:

5C4 = 5

Number of committees with only women:

8C4 = 70

Number of total committees:

13C4 = 715

Number of permitted comittees:

13C4 - 5C4 = 715 - 5 = 710

Number of comittees that aren’t all women (or at least one man)

710 - 8C4 = 710 - 70 = 640

7

Thanks to everyone for helping me clear this problem up; and yes this was a homework problem, but the homework is not collected and the answers are in the back of the book. Thanks again for the help.

8

I don’t understand how those formulas work. Can someone tell me what the Cs stand for? I’d love to know how to do problems like this.

9

It’s combinatorics. Basically, whenever you have a pool of stuff, and you want to know how many different combinations you can have, you use combinatorics.

If the order doesn’t matter (e.g. how many different ways can you pick 7 letters/numbers, where ABC1234 is the same as 1A2B3C4), it’s a combination, if the order DOES matter (e.g. how many different license plates can there be, since ABC1234 is different from 1A2B3C4), it’s a permutation.

The number of possible PERMUTATIONS of r objects from a collection of m objects is: [sub]m[/sub]P[sub]r[/sub]=(m!)/(m-r)!

Example: how many 3 letter words are there, using the 26 letters of the alphabet?

[sub]26[/sub]P[sub]3[/sub]=26!/23!=156,000
NOTE: this does not include words in which letters appear more than once, I think.

The number of possible COMBINATIONS of r objects from a collection of m objects is: [sub]m[/sub]C[sub]r[/sub]=(m!)/(r!((m-r)!))

Example: how many different combinations of three letters are there (abc is the same as cba)?

[sub]26[/sub]C[sub]3[/sub]=(26!)/(3!((26-3)!))=2600.

Aside from that, it’s just applying logic and basic probability to problems. For example, if I have a selection of 100 engines, and I know 5 are defective, if I pick 5 engines, I can calculate the probability that I selected no defective engines, exactly 3 defective engines, or even all five defective engines.

Example: What is the probability of selecting more than three defective engines?

From the pool of defective engines, I have selected three, four, or five. From the pool of non-defective engines, I have selected 2, 1, or 0, respectively.

There are [sub]5[/sub]C[sub]5[/sub] ways in which I can select 5 defective engines from a pool of five, and [sub]95[/sub]C[sub]0[/sub] ways in which I can select 0 defective engines from a pool of 95 non-defective engines. You multiply these answers together to get the total number of ways to select 5 defective engines, and you get 1.

There are [sub]5[/sub]C[sub]4[/sub] times [sub]95[/sub]C[sub]1[/sub] ways to pick exactly four defective engines and one good engine. This comes out to 475.

There are [sub]5[/sub]C[sub]3[/sub] times [sub]95[/sub]C[sub]2[/sub] ways to pick exactly three defective engines and two good engines. This comes out to 44,650.

Adding the three together, we get 45126 ways to pick three or more defective engines. There is a total of [sub]100[/sub]C[sub]5[/sub] choices, or 75,287,520 possible ways of picking five engines from a pool of 100, and 45126 of those involve picking three or more defective engines. Therefore, the probability of picking three or more defective engines from a pool of 100 engines, of which five are defective is 45126/75287520, or 5.99*10[sup]-4[/sup], or .0006, or .06%.

Additionally, the likelyhood of picking NONE of the five defective engines is ([sub]5[/sub]C[sub]0[/sub]*[sub]95[/sub]C[sub]5[/sub])/[sub]100[/sub]C[sub]5[/sub], or 57940519/75287520, or .76959, or 77%.

10

:smack: This should read: “What is the probability of selecting THREE OR MORE defective engines?”

Seriously, you have to be nitpicky about that sort of thing, else it changes your answer.

11

Thanks for the lengthy explanation Garfield226 but I still don’t get it. :frowning: How does (26!)/(3!((26-3)!)) =2600? I’m trying to multiply 3 by 23 then divide by 26 and other combos of how I think you might have got 2600 but I can’t come up with the right answer.

12

The !s are factorials.

3!, for instance, is 123, or 6.

26! is 12345678910111213141516171819202122232425*26.

13

How can I do 26! with a calculator?

14

If you don’t have a scientific calculator, you’ll just have to do it the long way, as I wrote it out in the last post.

15

How about with a scientific calculator?

16

Well, many of them will have a ! button. Just use that. The Windows scientific calculator has one.

17

if you don’t see ! then go into your probability function and it should be in there.

18

Oh, right. I was making a distinction between a scientific calculator and a programmable calculator, such as a TI-83.

If you have a TI-83+, enter the number, then push the MATH key, arrow over to PRB, and select number 4.

19

Many (scientific and graphing) calculators not only can do factorials, they can also do permutations and combinations. Look for keys labeled something like [sub]n[/sub]P[sub]r[/sub] and [sub]n[/sub]C[sub]r[/sub], or futz around with the menus, or if all else fails, RTFM. (Sadly, Windows’s pathetic excuse for a scientific calculator does not have these functions. Why is theirs so lame, anyway?)

Alternative method for calculating permutations and combinations:

To get [sub]n[/sub]P[sub]r[/sub], multiply together n * n-1 * n-2 etc. until you’ve multiplied r numbers altogether.
For example, [sub]26[/sub]P[sub]3[/sub] = 26 * 25 * 24. This works because you have 26 choices for what the first thing can be, 25 possibilities for the second, and 24 left to choose from for what comes third.

Then to calculate [sub]n[/sub]C[sub]r[/sub] (where order does not matter), divide [sub]n[/sub]P[sub]r[/sub] by r!
For example, [sub]26[/sub]C[sub]3[/sub] = 26 * 25 * 24 / (3 * 2 * 1).
This works because when you count permutations, you counted each combination 6 times, once for every possible order the three things could appear in.

20

I would think it would be.

c(5,3)c(8,1) is 54/2 * 8 = 80
280 +280 +80 + 640, right? Am I missing something???