When playing a game, you roll five regular 6-sided dice. How many

Kindly assist, GMAT Club Team.

Why is the answer to this question not 6^5 (since the outcome for each die is 6). Or is it because the order of the dice don’t matter?

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It is because order doesnt matter

Hi GMAT Club Team, kindly assist.

In the selection for Number 4, ”4 different numbers”: AABCD, why do we;
1. Lump up the selection for the single numbers (ie. 5C3), instead of 5C1* 4C1*3C1?
2. Also, why do we choose the number which appears twice first (ie. Why can’t we use 6C3*5C1 instead).

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Bunuel can you please provide a detailed solution to this question ?

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You can think of this problem as placing 6 different items into identical “bins” of different sizes. The only difference is that in some cases the bins may hold more items than they do in other cases. However, for each case the total number of items we ultimately place in the bins will be 5 out of the 6 total numbers.

We have the 6 items as 6 numbers:

#1, #2, #3, #4, #5, #6

Since the order doesn’t matter, we just need the number to “show up” in each case. What will matter, however, is how many times each chosen number shows up for each case.

Case 1: we can pick all 5 numbers as exactly the same.

Ex: 1-1-1-1-1

There are (6 c 1) ways to do so = 6 ways

OR

Case 2: 1-1-1-1 ……2

We can have 2 different numbers. However, for each possible combination of 2 numbers, we need to account for which “bin” the number will go in (the 4 item bin or the 1 item bin)

For instance: 3-3-3-3-4 ………is different from…… 4-3-3-3-3

This is true even though order does not matter.

(6 c 2) = 15 ways to get a combination of two numbers

2! = ways to shuffle the two numbers around for each of those 15 combinations

15 * 2 = 30 ways

OR

Case 3: 1-1-1….. 2-2

For this case, we again need to pick two different numbers out of 6.

And again, for each possible pair, we can shuffle the two numbers around in 2! Ways into the two different sized “bins”

(6 c 2) * 2! = 30 ways

OR

Case 4: 1-1-1 …4…..5

Now, for this case, we will pick 3 different numbers out of a possible 6.

6 c 3 = 20 ways

However, when we shuffle the numbers around in the second step we need to be careful. Because there are two bins with the same count (one number —- say 4, for example —- and another one number—- say 5, for example), we need to remove the overcounting.

If we just arranged in 3! Ways we would be including both cases such as:

1-1-1…4….5
and
1-1-1….5…4

This is similar to the anagrams when you have a word such as: SCOTT

The two T elements are identical, so you need to account for that fact.

We can arrange in: 3! / 2! = 3 ways

20 * 3 = 60 ways

OR

Case 5: 1-1……3-3…..5

Here, we are choosing 3 different numbers. So first we want to find the number of different ways to choose 3 numbers out of 6 possible numbers.

6 c 3 = 20 ways

Then, we again need to be careful when we arrange the three chosen numbers for each of the 20 ways.

If we were to just arrange the numbered in 3! Ways, we would be including the following cases because there is (again) two “bins” with an identical count.

3-3….4-4….5
and
4-4…..3-3….5

Since order does not matter, we only want to include one of these in the final count.

Can therefore arrange in:
3! / 2! = 3 ways

20 * 3 = 60 ways

OR

Case 6: 1-1…3…4…5

Here we are choosing 4 numbers out of the 6.

6 c 4 = 15 ways

This time, we have 3 “bins” with the same count of numbers (one and one and one)

If for each of the 15 ways we just moved the four numbers around in 4! Ways we would be over-counting and including cases such as the following:

1-1…3…4…5
1-1….3….5….4
1-1….4…..3….5
1-1…..4….5….3
etc.

And the order does not matter.
We can therefore arrange in: 4! / 3! = 4 ways

15 * 4 = 60 ways

OR

Case 7: 1-2-3-4-5

Finally, for case 7, we can have all different numbers show up for the 5 numbers.

Since order doesn’t matter, this is:

6 c 5 = 6 ways

Adding up cases 1 thru 7:

6 + 30 + 30 + 60 + 60 + 60 + 6 =

6 + (4) (60) + 6 =

252 ways

A

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Total outcome => 6^5=7776 (order is considered here)

But the question says order doesn’t matter so we have to unorder it.

So ways= 7776/(5!)=........

But that is not the answer.

KarishmaB Can u plz say why this method is not working ?

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This is a case of distributing 5 identical objects (5 dice) among 6 different groups such that a group may get no die. Think of it as 6 diff buckets with numbers from 1 to 6 written on them. We could place 2 identical dice in bucket 1, 1 die in bucket 2 and rest 2 dice in bucket 5. We get {1, 1, 2, 5, 5} on the 5 dice with no arrangement.
When identical objects are distributed among distinct groups, we use the partition method.

The identical objects are placed in a line and (n-1) identical partitions are used to create grouping. For 6 groups, we need 5 partitions.

All arrangements of DDDDD||||| will give us our possible outcomes.

|DD|D|||D means no die shows 1, 2 dice show 2, 1 die shows 3, no die shows 4 or 5 and 1 die shows 6.
There is no D in bucket 1, 2 Ds in bucket 2, 1 D in bucket 3, No Ds in buckets 4 and 5 and 1 D in bucket 6.

No. of arrangements of DDDDD||||| is 10!/5!*5! = 252

Note that 6^5 is not divisible by 5! so it will not give you the number of ways in any case.

6^5 is used when there are 6 ways to fill 5 distinct spots. Here the spots are not distinct. We cannot make them identical by dividing by 5!.
Division by 5! is used to un-arrange 5 distinct arranged objects. The spots may get the same number. To un-arrange 12345, we can use 5!. But what happens when we have 12233?

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