1. ## Number of ways

Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labelled as '1' and '2', the three blue balls are labelled '1', '2' and '3', and the five green balls are labelled '1', '2', '3', '4' and '5'. Find the number of ways of choosing 2 balls of identical colours.

5x4+3x2+2x1=28

but answer is 14... what is wrong with my workings?

2. ## Re: Number of ways

Originally Posted by Punch
Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labelled as '1' and '2', the three blue balls are labelled '1', '2' and '3', and the five green balls are labelled '1', '2', '3', '4' and '5'. Find the number of ways of choosing 2 balls of identical colours.

5x4+3x2+2x1=28

but answer is 14... what is wrong with my workings?
Does the order in which you pick the balls matter? In other words, say I wanted to know how many ways to pick two identical red balls, is choosing 1 then 2 considered a different way than choosing 2 then 1?

3. ## Re: Number of ways

Originally Posted by Prove It
Does the order in which you pick the balls matter? In other words, say I wanted to know how many ways to pick two identical red balls, is choosing 1 then 2 considered a different way than choosing 2 then 1?
Right! thanks

4. ## Re: Number of ways

Hello, Punch!

Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green.
The two red balls are labelled as '1' and '2', the three blue balls are labelled '1', '2' and '3',
and the five green balls are labelled '1', '2', '3', '4' and '5'.
Find the number of ways of choosing 2 balls of identical colours.

5x4 + 3x2 + 2x1 = 28 . permutations?

but answer is 14 ... what is wrong with my workings?

Prove It has a legitimate concern.

You have: . $\left(_5P_2\right) + \left(_3P_2\right) + \left(_2P_2) \:=\:20 + 6 + 2 \:=\:28$

Your approach is that choosing $(R_1,R_2)$ in that order
. . is different from $(R_2,R_1)$ in that order.

I would assume that the order is not considered.
The two above outcomes would be considered one outcome: $\{R_1,R_2\}$

The answer would be: . $\left(_2C_2\right) + \left(_3C_2\right) + \left(_5C_2) \:=\:1 + 3 + 10 \:=\:14$