Lesson: When to use Combinations

Comment on When to use Combinations

so what is the answer to the 50 students in school, electing Pres, VP, and treasurer? Do you mind explaining?
gmat-admin's picture

The answer is 117600. The solution for that question starts around 3:10 in the video

I have similar question to 8 people hand shake,

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

The outcome is not 56 but 28, I guess it's because of wording" each team plays with other team only once"
Is my reasoning correct?
gmat-admin's picture

There are two approaches we can use. One is mathematical, and the other involves some number sense.

Number sense: Once all of the teams have finished playing their games, ask each of the 8 teams, "How many teams did you play?" We'll find that EACH TEAM played 7 teams, which gives us a total of 56 games (since 8 x 7 = 56).

From here we need to recognize that every game has been counted TWICE. For example, if Team A and Team B play a game, then Team A counts it as a game, AND Team B also counts it as a game. Of course only one game occurred between these teams. So, to account for the duplication, we'll divide 56 by 2 to get 28.

The mathematical approach: there are 8 teams, and we want to determine how many different ways they can be matched to play a game. This is the same as asking, "In how many different ways can we select 2 teams to play a game?" Notice that the order in which we select the 2 teams does not matter. For example, choosing Team A then Team B to play a game against each other is the same as choosing Team B then Team A to play a game against each other. Since order does not matter, we can select 2 teams from 8 teams in 8C2 ways (= 28 ways)

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different centers is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter)
(A) 4
(B) 5
(C) 6
(D) 12
(E) 24
gmat-admin's picture

I have answered that question here: http://www.beatthegmat.com/distribution-centers-t263732.html

Please let me know if you have any questions

How did we arrive at the figure 15?
gmat-admin's picture

Good catch! I meant to say 12. I have edited my response accordingly (http://www.beatthegmat.com/distribution-centers-t263732.html)

Thanks!

How do combinations relate to the MISSISSIPPI rule, given that here the order of some letters matters, whereas the order in other letters (e.g. S and I) it does not matter?
gmat-admin's picture

I wouldn't place too much energy into trying to determine whether there's a relationship between combinations and the MISSISSIPPI rule. Only in very specific cases are they similar.

gmat-admin's picture

Allow me to expand on my comment above.

In the video question at https://www.gmatprepnow.com/module/gmat-counting/video/786 we note that the number of possible routes is equal to the number of ways to arrange the letters DDDRRRR.

In the video solution, we use the MISSISSIPPI rule to determine the number of ways to arrange the letters DDDRRRR.

HOWEVER, since there are only 2 different letters (R and D), we could have also used COMBINATIONS to determine the number of ways to arrange the letters DDDRRRR.

Here's what I mean:

One way to arrange the 7 letters (DDDRRRR) is to think of which letter will go in the FIRST spot, which letter will go in the SECOND spot, which letter will go in the THIRD spot, etc.

Of the seven possible spots, 3 of them must be D's and 4 must be R's.

So, let's first place ALL THREE D's.
To do this, we'll CHOOSE 3 of the 7 spots that will have D's
We can CHOOSE 3 spots from 7 spots on 7C3 ways (= 35 ways)
Once we've placed the 3 D's in those 3 spots, we'll place the 4 R's in the 4 remaining spots.
This will complete the arrangement.

So, the number of ways to arrange the letters DDDRRRR is equal to 7C3 (35)

You need to put your reindeer, Lancer, Prancer, Gloopin, and Bloopin, in a single-file line to pull your sleigh. However, Prancer and Lancer are fighting, so you have to keep them apart, or they won't fly.How many ways you can arrange your reindeer?
gmat-admin's picture

One approach is to first IGNORE the restriction about keeping Prancer and Lancer apart.

So, we can arrange the 4 reindeer in 4! ways (= 24 ways)

Of course, some of those 24 arrangements include cases in which Prancer and Lancer are TOGETHER. So, we must subtract those instances from 24.

To determine how many cases there are in which Prancer and Lancer are TOGETHER, let's "glue" them together to make ONE big reindeer (PRANCERLANCER).

So, we now have 3 objects to arrange: PRANCERLANCER, Gloopin, and Bloopin
We can arrange those 3 objects in 3! ways (= 6 ways).

Also note that we can also "glue" Prancer and Lancer together so that Lancer is first to get LANCERPRANCER, Gloopin and Bloopin.
Once again, we have 3 objects to arrange: LANCERPRANCER, Gloopin, and Bloopin.
We can arrange those 3 objects in 3! ways (= 6 ways).

So, the total number of arrangements where Prancer and Lancer are TOGETHER = 3! + 3! = 6 + 6 = 12

We'll now subtract that amount from the 4! arrangements that IGNORE the restriction about keeping Prancer and Lancer apart.

So, the TOTAL number of arrangements in which keeping Prancer and Lancer are APART = 24 - 12 = 12

Cheers,
Brent

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