5
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Alan Moreno de la Rosa Jan 26 '14 at 8:23
  • $\begingroup$ That is correct. $\endgroup$ – Davis Yoshida Jan 26 '14 at 9:17
3
$\begingroup$

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"!

$\endgroup$
2
$\begingroup$

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 $2250$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.