Question

Consider an Urn that contains:

→ 3 blue balls
→ 2 red balls
→ 5 green balls
Suppose we are drawing 3 balls from this urn, one after the other with no replacement. In how many ways can we draw EXACTLY 2 balls of the same colour?

`note = \\ no \: replacement \\ order \: matters`
Please don't answer randomly I need this​

Answer 1

I would argue that order does not matter, since e.g. drawing 1 red, then 1 blue, then 1 more red is the same as drawing 2 reds then a blue. Either way you've drawn exactly 2 red balls.

If we draw two blue, this can happen in

`\dbinom 31 \dbinom 21 \dbinom 10 \dbinom 71 = 42 \text{ ways}`

If we draw two red, this can happen in

`\dbinom 21 \dbinom 11 \dbinom 81 = 16 \text{ ways}`

If we draw two green, this can happen in

`\dbinom 51 \dbinom 41 \dbinom 30 \dbinom 51 = 100 \text{ ways}`

Then the total number of ways of drawing exactly 2 balls of the same colour is 42 + 16 + 100 = 158.

To see where these products come from, consider the blue case as an example. We can draw 1 blue frst, which leaves 2 more blues that can be drawn; then we draw 1 more blue, leaving 1 blue in the urn; then we don't draw the last blue, and instead draw 1 of the other 7 balls that are not blue.

If you insist on keeping track of the order in which the balls are drawn, we just need to multiply the total count by 3!, since that's the number of ways we can shuffle 3 balls around. In that case, 3! • 158 = 948.

If order matters but the same-coloured balls are identical, we would instead multiply by 3!/2! = 3, giving a total of 3 • 158 = 474.