In a certain mathematical activity, we have five cards with five diffe

Question

In a certain mathematical activity, we have five cards with five different prime numbers on them. We will distribute these five cards among three envelope: all could go in any envelope, or they could be broken up in any way among the envelopes. Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set. How many different sets can be produced by this process?

(A) 41
(B) 89
(C) 125
(D) 243
(E) 512

Kudos for a correct solution.

(A) 41
(B) 89
(C) 125
(D) 243
(E) 512

Bunuel · Accepted Answer

MAGOOSH OFFICIAL SOLUTION:

We have to consider different groupings. First, all the cards together.

Case I: all five cards in one envelope (2 envelopes empty) = one possibility

Now, two envelopes used, one empty.

Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities

Case III: three in one, two in another, one empty = 5C3 = 10 possibilities

Now, no empty envelopes, all three used:

Case IV: 3-1-1 split = 5C3 = 10 possibilities

Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.

N = 1 + 5 + 10 + 10 + 15 = 41

Answer = (A)

Shobhit7 · Answer

There are 5 Different Prime Numbers that needs to be split among three identical envelopes.
Possible split-ups for 5 Prime Numbers are as follows:
5-0-0: No. Of Sets possible: 5C5 = 1
4-1-0: No. Of Sets possible: 5C4 x 1C1 = 5 x 1= 5
3-2-0: No. Of Sets possible: 5C3 x 2C2 = 10 x 1 = 10
3-1-1: No. Of Sets possible: {5C3 x 2C1 x 1C1} / 2! ( 2! Is the correction for identical 1's)= 10
2-2-1: No. Of Sets possible: {5C2 x 3C2 x 1C1} / 2! ( 2! Is the correction for identical 2's)= 15

Counting total sets possible= 1 + 5 + 10 + 10 + 15 = 41.
Hence Answer Choice : A

sterling19 · Answer

I believe the answer is B.

Each of the five cards can go in 1 of 3 envelopes, so there are 3^5 possible combinations.
As we are using different prime numbers, the possible products will always be distinct from each other and there will be no overlap.

camlan1990 · Answer

Case 1: 1 used envelope => 1 way
Case 2: 2 used envelopes
- 4-1-0: Choose 4 from 5 cards: 5 ways
- 3-2-0: Choose 3 from 5 cards: 10 ways
Case 3: All envelopes used
- 3-1-1: Choose 3 from 5 and  no need to choose 1 from 2: 10 ways 
- 2-2-1: Choose 2 from 5 and choose 2 from 3, but two groups are the same => (10X3):2 = 15

Total: 1+5+10+10+15=41 => Answer: A

chetan2u · Answer

hi your total is 51 and not 41..
i am sure you would want to half the ways in case 3 for 3-1-1.. 20/2=10 ways as you have done for 2-2-1... 
you will then get 41...
but just a question, why are you taking the two groups as same..
they may have 2-2 prime numbers in each but remember the product would be different for each envelope everytime as no two different prime numbers can give you same answer when multiplied..

camlan1990 · Answer

Hi chetan2u,

I m sorry for this mistake. I highlighted my answer with RED above.

In case 3-1-1: After choosing 3 from 5, you dont need to choose 1 from 2. Because there is ONE way to put in order for two other cards.
Thanks

smartyguy · Answer

chetan2u

Hi chetan
Can you please help me in understanding this question as i am not able to grab it properly

Regards
SG

AkshdeepS · Answer

"Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set." . I could not understand these statements and how we used these statements in the solution. Very confusing..please help..

Bunuel · Answer

Say the 5 primes are: 2, 3, 5, 7, and 11.

These primes then are distributed in 3 envelopes. Say 2 and 3 are in envelope #1, 5 and 7 envelope #2 and 11 in envelope #3.

The product of primes in envelopes are ("numbers" of envelopes): 
For #1: 2*3 = 6;
For #2: 5*7 = 35;
For #3: 11.

So, set we get with this distribution is {6, 11, 35}. This is one of the sets possible.

The question asks  how many different sets can be produced ?

Hope it's clear.

chesstitans · Answer

I dont quite get it, what the prime number has to do with the question? I understand that those prime numbers should be factored into the solution.

guialain · Answer

This is a very tough one!

Product of the 5 different numbers 2*3*5*7*11=2310 (Prime numbers 2,3,5,7 and 11 are chosen just for example)

Possibility 1a: Envelop1 receive 0card - Envelop2 receive 0card - Envelop3 receive 5cards === 1-1-2310
Possibility 1b: Envelop1 receive 0card - Envelop2 receive 5card - Envelop3 receive 0card === 1-2310-1
Possibility 1c: Envelop1 receive 5cards - Envelop2 receive 0card - Envelop3 receive 0card === 2310-1-1

The 3 possiblities above will result in 1 set because we must put the envelop in order (lowest to highest).
in other words, we eliminate possibilities in which the numbers are not in increasing order.

So, solve this problem, you can asume that the first envelop has the lowest product and the third envelop has the highest product.

Similar logic tells you that there are 5 sets possible in the case: envelop1 has 0 card, envelop2 has 1 card and envelop3 has 4 cards
in fact, case 4-1-0 is not an acceptable set because any prime number is greater than 1. same logic, case 1-4-0 is not an acceptable set.
case 0-1-4 and case 0-4-1 toghether have 5 acceptable sets. Possibilites that are not accpetable in case 0-1-4 are acceptable in case 0-4-1 and vis-versa.

Same logic can be replicate to Case 0-2-3 and Case 0-3-2, which together have 10 acceptable sets.
Same for case 1-1-3, which has 10 acceptable sets.

The trick is then on case 1-2-2, in which half of the possibilities are not in ascending order and are not replaced in any other case.
So from the 30 possibilities (5C1*2C4), only 15 are in ascending order, therefore acceptable sets.

The total acceptable cases are 1+5+10+10+15, ie 41 sets

Mco100 · Answer

Is there a reason we don't need to choose 1 from 2

TheSpecialNone · Answer

Would it make any difference if the products of the cards exceed the number of cards ?
 Ex: we have the first 5 prime numbers - 2,3,5,7 and 11- and only 2 envelopes used 
For example 3-2-0 and 2,3,5 and 7,11 respectively inside, where the first product is smaller than the second.
Does it have any consequences in the problem ?

swap6098 · Answer

Yes exactly. They have not considered the case when product is less than a particular prime number.
For example if prime numbers are 2,3,5,7,101.
This kind of case has not been considered.
And even the solution given by Magoosh official sums to 51 not 41.

IanStewart · Answer

This is not a realistic GMAT problem, and its wording is deeply problematic. In math, a set is a collection of numbers that  is not in order . That's the definition of a set. If something is in order, it's a sequence. So a math question can never tell you to "put the three numbers in order, from lowest to highest, and that is our set", because you aren't making a "set" the instant you put things in order. All of the information about putting things in order is irrelevant here, but it naturally leads to the confusion expressed in some of the questions above, about what happens when one prime is much larger than the others.

I've also never seen a counting problem on the GMAT with anywhere close to as many cases as you need to consider when solving this one. I can't even recall an official GMAT problem where I needed to consider more than three cases, and those are very rare. Here, just identifying the cases requires some work, and then we end up with five of them (5/0/0, 4/1/0, 3/1/1, 3/2/0, 2/2/1), the last of which is a bit tricky to deal with (because you have to notice you need to divide by 2, since it doesn't matter in which order you pick the two pairs of primes).

The official solution is correct, if I'm correctly guessing what the question is trying to ask, but I wouldn't suggest any GMAT test takers be concerned if they either find the wording confusing or find the question difficult or time-consuming, since you won't see a question like this on the real test.

Saby1098 · Answer

1. Try the grouping according to 
a. All in one envelope = 1 
b. 4 in one envelope, 1 in another, 0 in another = Number of ways = 5C4 = 5 
c. 3 in one envelope, 2 in another, 0 in another = 5C3 =10 
d. 3 in one envelope, 1 in another, 1 in another = (5C3 * 2C1 ) / 2! = 10 
e. 2 in one envelope, 2 in another, 1 in another = (5C2 * 3C2)/2! = 15 
f. Add all and since we are calculating the product, the permutation or order doesn't matter

Bunuel  - If the question didn't ask the product to be calculated to determine the set, would the answer be (40) * 3! + 1? 
Please let me know your thoughts on this

karantambat · Answer

does gmat actually have questions of this difficulty? or these are just for practice?

Fdambro294 · Answer

So if I'm understanding the question correctly....

Because the Ordering of the "Set" will be chosen based on the Numbers placed in the envelope, we do NOT have to Arrange the Groups. In effect we have 3 "Identical Groups."

This becomes a Question the same as: "How many ways are there to Distribute 5 Distinct Items ------> to 3 Identical Groups, in which the Ordering of the Groups does NOT Matter?"

All that matter is WHICH of the 5 Distinct Items are grouped together and which are NOT grouped together in "Stacks", so to speak.

Case 1: 5 - 0 - 0 
1 Way

Case 2: 4 - 1 - 0
We need to choose which 4 of the 5 Distinct Primes will be grouped together in 1 of the "Identical" Envelopes. 
For each Unique Distribution, once the 4 are chosen, there will be 1 remaining that will be placed in an envelope. 
the Last envelope will have NONE and get the Number "1".
The Set will then automatically be chosen based on our "Stacks" of Prime Numbers

"5 choose 4" ------> 5! / (4!)(1!) = 5 Ways

Case 3: 3 - 2 - 0
We need to choose which 3 of the 5 Distinct Primes will be grouped together in 1 of the "Identical" Envelopes. Once we choose the 3, automatically the 2 will be placed in the 2nd envelope.

"5 Choose 3" -----> 5! / (3!)(2!) = 10 Ways

Case 4: 3 - 1 - 1
Which of the 3 Distinct Primes will be grouped together in 1 Envelope. The other 1 and 1 will automatically be chosen once we choose the 3 for each Unique Distribution and our "Set" will be made.

"5 choose 3" -----> 10 ways

Case 5: 2 - 2 - 1
We will put 2 Primes in 1 Envelope. Then another 2 Primes in a 2nd Envelope.

However, if we just use the Normal Combination Formula (one after another) we will end up over counting, because the Size of the Groupings is the SAME (2 in one envelope and 2 in another envelope).

For Each 2 Distributions we will account for, only 1 will be a Unique Distribution that we want to keep. Therefore, in order to remove the over-counting, we need to Divide by 2!

"5 choose 2" and "3 choose 2" ------> 5! / (3!)(2!) * 3! / (2!)(1!) = 5! / (2!) (2!)

then Divide by 2!

5! / (2!) (2!) * (1 / 2!) = 15 ways

Counting up each scenario:

1 + 5 + 10 + 10 +15 = 41 Ways

-A-

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