fiza gupta wrote:Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?
A)30
B)60
C)120
D)240
E)480

OA:B

Tricky (700+) level question!

Since half of the marbles are red, there are only two possible arrangements for the RED marbles:
1) R _ R _ R _ R _ R _ R _
2) _ R _ R _ R _ R _ R _R
These are the only 2 ways to adhere to the rule that no two adjacent marbles are of the same color.

So, at this point, our objective is to arrange the 2 blue, 2 green and 1 yellow marbles in the spaces.
To do this, I'm going to ignore the red marbles and focus solely on the 2 blue, 2 green and 1 yellow marbles.

In how many ways can we arrange 2 blue, 2 green and 1 yellow marbles?
-------------------------------------------------
ASIDE: When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-------------------------------------------------
Now back to arranging our 2 blue, 2 green and 1 yellow marbles.
There are 5 letters in total
There are 2 identical B's (blue marbles)
There are 2 identical G's (green marbles)
The total number of possible arrangements = 5!/2!2! = (5)(4)(3)(2)(1)/(2)(1)(2)(1) = 30

Notice for EACH of these 30 arrangements of 2 blue, 2 green and 1 yellow marbles, there are 2 ways to add in the 5 red marbles (as discussed earlier).

Here's what I mean:
One way to arrange 2 blue, 2 green and 1 yellow marbles is as follows: BGBYG
There are 2 ways to add the 5 red marbles to this arrangement:
1) RBRGRBRYRG
2) BRGRBRYRGR

Another way to arrange 2 blue, 2 green and 1 yellow marbles is as follows: BBGYG
There are 2 ways to add the 5 red marbles to this arrangement:
1) RBRBRGRYRG
2) BRBRGRYRGR

And so on....

As we can see, for EACH of the 30 arrangements of 2 blue, 2 green and 1 yellow marbles, there are 2 ways to add in the 5 red marbles.

So, the TOTAL number of arrangements of the 10 marbles = (2)(30)
= 60
= B

fiza gupta wrote:Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?

A)30
B)60
C)120
D)240
E)480

OA:B

Great solutions. Nothing to add except that we need to decide which color marble to first arrange.

We see that there are 5 red colored marbles, and 5 other three color marbles.

Since the placement of red marbles, leaving one place before and after each of them will help accommodate other color marbles, we must select red marbles to arrange first.

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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