Of the 12 temporary employees in a certain company, 4 will

Select three women out of five in 5C3=10 ways.Select one man in 7 ways or 10x7=70 different groups of 3 women and 1 man.

Possible groups = Possible groups of 3 women * possible groups of 1 men

Since order does not matter, its a combination problem.

=[5!/(3!2!)} * [7!/(6!1!)]
=70

Select 3 out of 5 = 5c3 = 5
Select 1 out of 7 = 7c1 = 7
Total = 5c3 x 7c1 = 5x7 = 35

another approach :

one scenario : [WWWM]

now we have 5 ways for the first women , 4 ways for the 2nd women and , 3 ways for the third women.

and 7 ways to select one man

so 5*4*3*7 = 420

now the three women are same so combinations with win them are irrelevant the 3 women can we arranged in 3! ways

hence 420/3!= 420/6= 70

You need to decide if the problem at hand asks for permutation (order matters) or combination (order not important) or hybrid of the two.

In your first problem you have two separate slots for males and females. The problem doesn't care about the order of the selected employees so each slot contains a combination and you multiply them: 5C3 * 7C1, to get the total number of possibilities.

In the probability problems of coin or die tosses, you always calculate the probability of a particular outcome having a fixed order = permutation. The reason for that is that the formula for multiplying probabilities is defined only for events in a logical sequence: first even A, then event B whose probability may depend on A, etc. If the problem doesn't care about the order at the end, which is usually the case, it is asking for a the probability of a COMBINATION, so you have to find how many of the permutations merge into the desired combination and sum up their probabilities because they are exclusive (cannot happen at the same time).

Note that in coin or die tosses you always pretend they are tossed in a sequence one after another (otherwise you can't apply the formula for multiplying probabilities) even when they are actually tossed at the same time.

Thanks for replying, tutorphd. What do we do we have to deal with conditional probability?

Let's say you have this question -

A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

A, 3/40000, B 1/3600, C. 9/2000, D, 1/60, E, 1/15

In this question

Why won't we add the probability of two cases - (Case I - Junior first then Senior) + (Case II Senior First then Junior)

= 60/1000 * 1/800 + 60/800 * 1/1000

What is wrong with this ?

Answer is A

Wouldn't JS , SJ be two muutually exlusive possibilities, and we should apply the probabiliy rule P(A or B) = P(A) +P(B) - P(A and B)

Given this rule shouldn't we be doing the above proposed solution, however the answer I get is double of the official answer?

I kind of understand the point that you made in your post above, as a request can I please ask you to give me an example of the points that you made in your previous post. I will really appreciate if you can give me some examples to explain your previous explaination. I am doing fine in combinatorics at this point but this concept of multiplying with number of things in the end is messing me up often. Please help.

The general formula for multiplying probabilities is P(A and B) = P(A)*P(B|A) and uses the conditional probability P(B|A). For independent events, the conditional probability P(B|A)=P(B) and the formula simplifies to P(A and B) = P(A)*P(B), it is still called "conditional probability formula". The ordering in this formula is arbitrary logical ordering (first event A, then event B) obtained by tracing a probability tree. It has nothing to do with ordering in time with which you are mixing it up.

In the problems with coin or die tosses, you can label each coin, say 'coin A', 'coin B' ... and trace a probability tree by first considering the outcome for coin A, then coin B, coin C... You DON'T add that to another probability tree by first considering the outcome of coin B, then A, then C ... You use a SINGLE probability tree = single arbitrary logical ordering which has nothing to do with which coin was tossed first or second in time. That is why problems in which you toss coins/dice at the same time are equivalent to problems where you toss then in sequence, because the order is arbitrary logical ordering BY LABELING THEM and tracing a SINGLE probability tree, not by ordering them in time.

In the sibling pair problem, the labels are already there 'junior' and 'senior'. You trace the probability tree either as outcomes for junior first and senior second, or the the other way around BUT NOT BOTH. You are getting a double result because you are trying to add the results of two probability trees, not two 'exclusive events'.

In the coin toss problems, you are tracing a single tree with an arbitrary logical ordering you have chosen, say outcome of 'coin A' first, outcome of 'coin B' second etc. The exclusive events/outcomes are separate branches in that same tree, say {A,B}={H, T} or {T,H}, not part of another tree.

Number of ways to select 3 women out of 5 is \(5C3\) and the number of ways to select 1 man out of 7 is \(7C1\).
Therefore number of possible groups of 4 temporary employees consist of 3 women and 1 man are \(7C1 * 5C3 = 70\).
There are 70 possible groups.
+1D

Combinations will do.

5 out of the 12 employees are women, 3 of whom will be chosen and 2 whom will not. Therefore: The women = \(\frac{5!}{3!2!} = 5*2 = 10\)
7 out of the 12 employees are men, 1 of whom will be chosen en 6 whom will not. Therefore: The men = \(\frac{7!}{6!1!} = 7\)

\(7 * 10 = 70\) possible groups consisting of 3 women en 1 man.

Take the task of selecting the employees and break it into stages.

Stage 1: Select the 3 women
The order in which we select the women does not matter, so we can use combinations.
We can select 3 women from 5 women in 5C3 ways (= 10 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select the 1 man
There are 7 men, so we can complete this stage in 7 ways.

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus select the permanent employees) in (10)(7) ways ([spoiler]= 70 ways[/spoiler])

Answer: D

Note: the FCP can be used to solve the majority of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

Question - Why aren't we multiplying 70 with 4 to count possibilities to arrange 3w and 1m in the group, i.e. mwww, wmww, wwmw, and wwwm? I guess I am confusing/missing something here.

Thanks!

Those 4 cases are all the same.
For example, the group consisting of Ann, Bob, Bill, and Bart is the same as the group consisting of Bob, Bart, Ann, and Bill

Cheers,
Brent

We are asked to find the number of ways of choosing 3 women from a group of 5 women and 1 man from a group of 7 men.

Let’s first find the number of ways one can choose 3 women from 5. Since the order of how the 3 women are chosen doesn’t matter, we use combinations:

5C3 = 5!/(3! x 2!) = (5 x 4 x 3)/3! = 60/6 = 10

Similarly, the number of ways one can choose 1 man from a group of 7 men is:

7C1 = 7!/(1! x 6!) = 7

Finally, the number of ways we can choose 3 women from 5 women and 1 man from 7 men is:

5C3 x 7C1 = 10 x 7 = 70

Answer: D

Hi VeritasKarishma Bunuel chetan2u

Why have we divided by 3! here instead of by 4!/3! ? Isn't the number of ways WWWM can be arranged in 4!/3! ?

Thanks in advance.

12 temporary employees [As 5 are women and 7 will be men]

=> 4 employees consist of 3 women and one man: \(^5{C_3} * ^7{C_1} = 10 * 7 = 70\)

Answer D

You have taken WWWM, but order does not matter while choosing W.
5*4*3 means we have taken order in consideration, that is it is permutation 5P3. To convert this back to combinations where order does not matter, that is 5C3, we require to divide by the way these 3 Ws can be arranged within themselves and that is 3!.

5P3=5!/(5-3)!
5C3=5!/3!(5-3)!.......division by 3! is over and above 5P3.

Thanks for the reply.

Order doesn't matter for the whole WWWM arrangement right ? If we were told to find different arrangements of WWWM it would have been 4!/3! right ? Then why is in this case the arrangements just 3! ? It's as if M can't change the position.