There are x people and y chairs in a room where x and y are

@ Dmitry Faber..One of my pain points in PnC is when the number of things to be allocated is more than the number of people...
Number of ways to distribute
1. 100 identical pencils among 10 different students..Each student can carryonly one pencil (As is in the case of chair above)
--> 1 way

2. 100 different pencils among 10 different students..Each student can carry only one pencil
--> 100C10*10!

3. 100 different pencils among 10 Identical students..Each student can carry only one pencil
--> 100C10

4. 100 Identicalpencils among 10 different students..Each student can carry any number of pencils At least 0 & Max 100
-->

5. 100 different pencils among 10 different students..Each student can carry any number of pencils At least 0 & Max 100
-->

I used the same logic but I am stuck...I think 4 & 5 have to be approached case wise..
For eg...1 student receives all pencils..2 students receive all pencils..till 10 students receive all pencils

Am I correct in my logic?Experts please chime in...

Dear JusTLucK04,
I'm happy to respond.

Understand that these last two questions, #4 and #5, are leagues beyond anything the GMAT could possibly ask. We are getting here into contest-level math that few people in the population could solve. These make 800+ level questions look easy.

Here's how I would think about #4 --- think about 100 pencils in a row, with spaces between them, and spaces at each end:
_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P_P .....
That's 101 possible space. We are going to put nine "lines" in nine random spaces
_P_P | P_P_P_P_P_P_P_P || P_P_P_P_P_P_P_P_P_P_P_P_P_P_P | P .....
The nine lines will demarcate the shares of pencils of the ten individuals. Reading from left to right on this particular spread, individual #1 gets 2 pencils, individual #2 gets 8 pencils, individual #3 gets zero pencils (no pencils between the second & third lines), individual #4 gets 15 pencils, etc. etc.
As here, two or more lines could land in the same space: this would simply indicate that some individual or individuals get zero pencils. What would be tricky about this is: exactly how many different ways can the nine lines land in those spaces? It will be tricky to eliminate double-counting. One might have to count this with case, but then we are only counting the cases of nine things: still, that's 30 cases to consider. (I know it's 30 from the Partition Function.)

Then, for #5, if all 100 pencils are different, then everything we did for #4 we would have to repeat for each order of the 100 different pencils. Thus, whatever the answer to #4 is, we would multiply it by 100!, which is approximately 9.332621545 x 10^157. That's a number much more than all the atoms & subatomic particles in the Visible Universe. That's big.

Does this help?
Mike

Responding to a pm:

First I would like to point out that there really isn't anything called "identical students". Even twins have different thumb impressions so they aren't identical, strictly speaking. Instead you could say "10 identical baskets" etc.

Coming to case 4, this is a case of identical objects among distinct groups. I have discussed two methods to solve this one using "5 identical objects in 4 different groups" in the post below. First method uses cases and second method uses partitioning. The first method will be too cumbersome if the number of objects is huge. So focus on the second method.


Case 5 is the easiest. When distributing distinct things in distinct groups, there are actually no complications. The assumption is that you want all pencils to be distributed. In that case each pencil can be given away in 10 ways (to any of the 10 students). So total number of ways of distributing the 100 pencils will be
10*10*10........ (100 times) = \(10^{100}\)
This is question 1 in the link I gave you. 5 different fruits among 4 different students.

Now think of a problem where it is acceptable to not distribute all pencils and instead store some/all of them away. In how many ways can you do that?

(1) x + y = 12
if X=5 , then Y=7. more chairs less children . we have to choose 5 chairs out of 7 . total number of ways . \(^7C_5\) .
now 5 children can be arranged in 5!ways .
Total number of ways = \(^7C_5 * 5!\)

if X=7 , then Y=5. more children less chairs . we have to choose 5 children out of 7 . total number of ways . \(^7C_5\) .
now 5 children can be arranged in 5!ways .
Total number of ways = \(^7C_5 * 5!\)

A is sufficient.

(2) There are more chairs than children.
Clearly insufficient.

VERITAS PREP OFFICIAL SOLUTION:

There are x children and y chairs.
x and y are prime numbers.

Statement 1: x + y = 12

Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:

Case 1: x=5 and y=7
There are 5 children and 7 chairs.

Case 2: x=7 and y=5
There are 7 children and 5 chairs

At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.

Case 1: x=5 and y=7
If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!

Case 2: x = 7 and y = 5
If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.
Total number of arrangements would be 7C5 * 5!

Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.

So actually this statement alone is sufficient! Most people would not have seen that coming!

Statement 2: There are more chairs than people.
We don’t know how many children or chairs there are. This statement alone is not sufficient.

Answer: A.

We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!

I have a question on this approach. Given that we don't know the difficulty level of each question on the real test, how can we tell what a C-trap is? I've seen easy questions where I've suspected a C-trap, but it turned out to just be an easy question!

A wider question linking into this is that I'm struggling with the 'get the easy / medium ones right' approach to the GMAT - sometimes I can spend a minute or two on a problem before being able to gauge that it is too difficult for me. Any suggestions on knowing when to skip a question because it's too hard?

You can't say whether it is an "easy C" trap or just an easy question with answer C. But there are some pointers that can help:

Other than the first few questions, you are unlikely to find questions "easy" in GMAT. Its algorithm adapts to you so once it approximates your level, it will give you questions around it. You will not find the questions around your level easy. So if the 30th question seems just too "in your face" (C) to me, I will back up to ensure that no one statement alone is sufficient. Of course, you could get an easy question later on in case the software wants to ensure you see questions from all different subjects (one of the many constraints which help it decide which question to give you next) but even if you back up and find nothing sneaky, you will know that it is just one of those "easy questions with answer (C)"

What helps is having very strong concepts. It makes sense that combinatorics will not distinguish between animate and inanimate objects. If I have 7 distinct objects on one side and 5 distinct objects on the other side and I have to make pairs of two objects - one from each side, I can do it in say, X ways. Does it matter which side has animate objects and which has inanimate? Not at all!

You need to read the question to judge whether it seems easy or hard. That takes about 30 secs. If it seems hard and is from an area of weakness, you can choose to skip it if you know that you are usually short on time. If it is from an area of strength, you will probably be able to get it if you invest a couple of mins in it. Your goal should be to complete every question. Thinking of skipping only if you are usually short on time.

What if there are 10 chairs and 2 people or 9 chairs and 3 people or 8 chairs and 4 people? will that not change the number of ways ? Since it is not specified that the number of people should be greater than 5 these 3 scenarios are possible.

Krenuka Watch out--you're missing a constraint! Both x and y must be prime. That's why they can only be 5 or 7.

The only problem with this question is that it assumes we won’t count blank chairs as part of the order…OG would never ask a question like this.

rikinmathur

That's actually not a problem. The question never says anything about placing the chairs in order. We are just assigning people to chairs, just like we might assign gifts to recipients or drivers to trains. The result we find will be all the different ways to assign 5 people to 7 chairs. By default, this will specify which two chairs are blank. There's nothing else to consider, since we aren't also asking which chair is 1st, 2nd, etc.

Alternatively, you could imagine the 7 chairs in a row from 1 to 7, and we are saying which ones are blank. In that case, we would see an order. The math is the same either way.

It is given that only one person can sit in a chair, but it is not given that number of chairs is more than number of people...then how it is A?

If we combine 1 & 2 then we get the answer... please correct me if I'm wrong

Posted from my mobile device

If 5 chairs and 7 people then it is only 120...right?

Posted from my mobile device

____________________________
Please re-read the highlighted part!

No--if that were the case, then everything all of us said above, including the official explanation, would be wrong. That's not impossible, but generally when it seems that way to me, *I* have made an error. Don't assume otherwise!

As shown above, it really doesn't matter whether we have 7 people in 5 chairs or 5 people in 7 chairs. Here's another way to think about it:

5 people in 7 chairs:

The first person can be in any of the 7 chairs, the next in any of the remaining 6, etc. So that's 7*6*5*4*3

7 people in 5 chairs:

The first person can be in any of the 5 chairs, or they can be one of the two people who stands, so there are 7 options for them. The next person has 6 options, etc.
But the two people who stand aren't in distinct positions from another, so we divide that by 2. That's 7*6*5*4*3*2*1/2 = 7*6*5*4*3.

It really is the same answer either way! That's what makes this tricky.