# Having trouble with a combinatorics question.

I'm not so good at combinatorics, but I want to know if my answer for this question is right. Originally this question is written in spanish and it says:

Se dispone de una colección de 30 pelotas divididas en 5 tamaños distintos y 6 colores diferentes de tal manera que en cada tamaño hay los seis colores.¿Cuántas colecciones de 4 pelotas tienen exactamente 2 pares de pelotas del mismo tamaño (que no sean las 4 del mismo tamaño)?.

And here's a translation made by me:

A collection of 30 balls is available, separated in 5 different sizes and 6 different colors in a way that in each size there are six colors. How many collections of 4 balls have exactly 2 pairs of balls of same size (which those 4 balls aren't of same size)?

I first wrote this table:

\begin{array}{c|c|c|c|c} \cdot & Size 1 & Size 2 & Size 3 & Size 4 & Size 5 \\ \hline Color 1 & ① & ❶ & ⒈ & ⑴ & ⓵ \\ Color 2 & ② & ❷ & ⒉ & ⑵ & ⓶ \\ Color 3 & ③ & ❸ & ⒊ & ⑶ & ⓷ \\ Color 4 & ④ & ❹ & ⒋ & ⑷ & ⓸ \\ Color 5 & ⑤ & ❺ & ⒌ & ⑸ & ⓹ \\ Color 6 & ⑥ & ❻ & ⒍ & ⑹ & ⓺ \end{array}

So, an example of a collection of 4 balls that have exactly 2 pairs of balls of the same size is:

①②❶❷

So, for the first column (Size 1) there are 15 combinations of having 2 balls:

①②  ②③  ③④  ④⑤  ⑤⑥
①③  ②④  ③⑤  ④⑥
①④  ②⑤  ③⑥
①⑤  ②⑥
①⑥


Which is the same as:

$$C_{6}^{2} = \frac{6!}{(6-2)!2!} = 15$$

Or the same as:

$$\sum_{k=1}^{5}k = 15$$

Then, for each row we have 10 combinations:

①❶   ❶⒈  ⒈⑴  ⑴⓵
①⒈  ❶⑴  ⒈⓵
①⑴  ❶⓵
①⓵


Which is the same as:

$$C_{5}^{2} = \frac{5!}{(5-2)!2!} = 10$$

Or the same as:

$$\sum_{k=1}^{4}k = 10$$

And so, by the rule of product I say that the number of collections of 4 balls having exactly 2 pairs of the same size is:

$$C_{6}^{2} C_{5}^{2} = 150$$

I'll be grateful if someone check my answer and give me further details :D

Not quite correct. Here's a breakdown of the choices you need to make.

1. You will need to choose two sizes of ball, so do so: $\binom{5}{2}$
2. For the first size, choose two colors: $\binom{6}{2}$
3. For the second size, choose two colors: $\binom{6}{2}$

Combine these to count your total.

• So, for instance, if the question was "How many collections of 4 balls have exactly 2 pairs of same color?" It would be: 1.Choose two colors $\binom{6}{2}$ 2.for the first color select one size $\binom{5}{2}$ 3. for the second color choose one size $\binom{5}{2}$ So there will be 1500 collections. Is that correct or am I wrong? – Alan Moreno de la Rosa Jan 26 '14 at 8:23
• That is correct. – Davis Yoshida Jan 26 '14 at 9:17

In this kind of problems, the key idea is to transform it into steps and cases in a way that you can use "Addition law" and "Multiplication law". Here you need tow different pairs of balls of the same size, so

Step 1: Select 2 different sizes from those existing 5 sizes;

Step 2: Select 2 balls in each of those 2 selected sizes;

Step 3: Use multiplication law.

And don't forget that "select" means "combination"!

There are $\binom{5}{2} = 10$ ways to choose two distinct sizes. For each of those ways, there are $\binom{6}{2} = 15$ ways to choose two distinct colors for the first pair, and another $\binom{6}{2}$ ways to choose two distinct colors for the second pair. So your answer is $2250$.