In how many ways can n different balls be arranged in r slots?

Question

(1) Given r balls and n slots to arrange the balls in, the balls can be arranged in 210 ways.

(2) If we only need to select and not arrange n balls for r slots, we can do it in 35 ways.

Source:  Nova GMAT

Dewanshi · Answer

can someone provide explanation for the question.. thanks!!

pulak1988 · Answer

explaination please.

Sajjad1994 · Answer

Dewanshi Official Explanation OA.jpg Answer: A

Commendlionel · Answer

Hi, I still don't quite understand the OE.

For statement 1, why is nPr still being used when the situation is swapped to r balls arranged in n slots? Shouldn't it be rPn now? And if so, how did the answer derive nPr?

DexterZabula · Answer

They need to mention which is greater - n or r?

Nups1324 · Answer

Hi, I did not understand the attachment by Sajjad. Is there any alternative way to solve this? Bunuel chetan2u IanStewart ScottTargetTestPrep yashikaaggarwal VeritasKarishma BrentGMATPrepNow GMATinsight Thank you

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chetan2u · Answer

A similar question but a better wording have following solution. Lot of issues in the question

In how many ways can n different balls be arranged in r slots?
This means we are looking for r*(r-1)*(r-2)*....(r-(n-1)).
Reason : The first ball can be put in any of the r slots, the next one in remaining r-1, and the last one or nth one in r-n

(1) Given r balls and n slots to arrange the balls in, the balls can be arranged in 210 ways.
This means we are given  n*(n-1)*(n-2)*....(n-3)=210 .
Reason : The first ball can be put in any of the n slots, the next one in remaining n-1, and the last one or rth one in n-r.
Let us factorize 210 such that it becomes product of consecutive integer.
210=2*3*5*7=7*6*5=7(7-1)(7-2)=n(n-1)..(n-(r-1))
So n=7 and r-1=2 or r=3
Now we can find 7 balls in 3 slots. => 3*2*1
The value of balls is more than the number of slots, so we cannot arrange all 7 balls in 3 slots!!
But sufficient to get value of n and r.

(2) If we only need to select and not arrange n balls for r slots, we can do it in 35 ways.
When we are selecting something we can have two values as nCr=nC(n-r)
Here 7C3=7*6*5/(3*2)=35, but also 7C4=35
So 7 balls in 3 OR 4 slots.
Insuff

A

Edit : I would agree to Ian that it is a poorly written question but I have just given a solution that may fit in in other questions.

IanStewart · Answer

No, because there's no way to even guess what the question means. How many balls can go in one slot? Reading their solution, we're meant to assume only one ball can go into a slot, but there's no way to guess that from the wording of the question (and I'd assume the opposite, just because if you see this kind of wording in any combinatorics book, it's almost always a "partition problem" where more than one ball can go in a slot or box). And then if, as the question apparently wants us to assume, each slot can only contain one ball, what is the answer supposed to be to the question "in how many ways can 5 balls be arranged in 2 slots?" The only reasonable answer to that question is "zero", because it's impossible to arrange 5 balls in 2 slots; there aren't enough slots. But they seem to want the answer to that question to be (5)(4). And, reading Statement 2, what is the answer to the question "in how many ways can you select and not arrange 2 balls for 5 slots?" There's presumably only one way to do that -- you'll use all of the balls. They seem to think, from the solution, that the answer is (5)(4)/2! = 10, but that's the number of ways to choose the slots, not the balls. Even when n > r, it's not clear what Statement 2 even means.

For the question to even make sense, it would need a lot of additional words to clarify the meaning, and to clarify how to answer a question when n > r and when r < n. Even if, by pure exercise of psychic ability, you guess what the question is trying to say, it is absurdly complicated to analyze Statement 2 the way that solution does. If Statement 2 is true, we might have 35 balls and 1 slot. Then the answer to the original question is 35. But we know, since DS Statements are consistent, that for some other numbers the answer to the question is 210, since it must be possible for Statement 1 to be true when we use Statement 2 alone. So Statement 2 cannot be sufficient. You certainly don't need to use the fact that nCr = nC(n-r) or even notice that 7C4 = 7C3 = 35.

Anyway, this question is far more confusing than helpful, and is nothing like what you see on the GMAT. I'd recommend instead working from questions that are properly worded and within the scope of the test.

ScottTargetTestPrep · Answer

One way of making sense of this question is to consider the question for specific values of r and n. For instance, if r = 3 and n = 5, the question becomes "In how many ways can 5 different balls be arranged in 3 slots?". Assuming no slot can be empty and no more than one ball can go in each slot, we need to choose three balls and arrange them, which can be done in 5P3 = 5 * 4 ways. This shows us that when n > r, the answer to the question is nPr.

Let's also consider the same question when r > n. If r = 5 and n = 3, then we are arranging 5 balls in 3 slots. In this scenario, assuming again only one ball per slot and no empty slots, we must now choose 3 balls and arrange them, which can also be done in 5P3 ways. Thus, when r > n, the answer is rPn.
Finally, if r = n, the answer is n! = r!.

Now, we can analyze the statements. Statement one alone tells us that rPn = 210 if r > n and nPr = 210 if n > r. Recall that the answer to the question was also rPn if r > n and nPr if n > r, so this statement tells us exactly what we need to know. Notice that assuming this statement, we can conclude that n cannot equal r since n! = 210 is not satisfied for any positive integer.

For statement two, we should observe that the minimum value of n which can satisfy nCr = 35 is 7 because nCr cannot have a factor of 7 for the values of n which are less than 7. Trying n = 7, we see that 7C3 = 7C4 = 35. Thus, r can be 3 or 4. If r = 3, then the answer to the question is 7P3 = 7 x 6 x 5 and if r = 4, then the answer is 7P4 = 7 x 6 x 5 x 4. Since we have multiple answers depending on the value of r, this statement is not sufficient to answer the question.

Answer: A

In how many ways can n different balls be arranged in r slots?

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