PnC Problem!

chaitanyabhansali wrote:Please solve this for me with a lucid step-by-step solution!

question: In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

I like Brent's solution method, but here's an alternative to solving this "internal order" type question:


1) Assume an easier state - all shirts are different colors (e.g. red, blue, purple, green, white, yellow).
for such a case, we will have 6!=6*5*4*3*2*1 different ways of distributing the shirts:
The first kid has 6 shirts to choose from.
the second kid chooses a shit from the 5 remaining.
4 for the third kid..
etc.

2) Phase 2: make the transition from six colors to only two colors: divide by the size of each group factorial. In this case, 3 green shirts and 3 red shirts means we divide the original 6! by 3! and 3!, resulting in 6!/3!3!.

chaitanyabhansali wrote:Please solve this for me with a lucid step-by-step solution!

question: In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

The reason we divide by 3! twice is that the internal order within each group does not matter, so we need to discount duplicate arrangements.

When arranging 6 different colors, we're counting these arrangements as different:

R, B, P, G, W, Y
R, B, P, G, Y, W
R, B, P, Y, G, W

etc.

But if we transition to two "groups", i.e replace B and P with red shirts (R) and W and Y with Green shirts (G), then all three arrangements above are actually only a single arrangement:

R, R, R, G, G, G
R, R, R, G, G, G
R, R, R, G, G, G

For a group of 3 (green, white, yellow shirts, for example), there are 3! different ways of arranging the internal order of the members in the group, all of which are acutally a single option when moving to G, G, G. Therefore, divide the original 6! by 3! to discount the duplicates of the green group, and 3! to discount the duplicates of the red group.

If the same problem had 2 identical green shirts, 2 identical red shirts, and 2 identical blue shirts, then the number of ways of distributing the shirts to six kids would be?

If the same problem had 2 identical green shirts, 2 identical red shirts, and 2 identical blue shirts, then the number of ways of distributing the shirts to six kids would be?

the answer is 90. Thanks for the explanation provided.

In GMAT, what would be the difficulty level of this question?

aravindan_v wrote:

If the same problem had 2 identical green shirts, 2 identical red shirts, and 2 identical blue shirts, then the number of ways of distributing the shirts to six kids would be?

the answer is 90. Thanks for the explanation provided.

In GMAT, what would be the difficulty level of this question?

High - I'd say only a 49+ student will encounter such a thing. Internal ordering questions are rare.