# What is the intuition behind using binomial coefficient in the problem?

A box has 10 red coloured and 15 blue coloureds pens. If you were to pick out 7 pens (without replacement), what is the probability that you get exactly 2 red coloured pens? The answer is $$\frac{{10 \choose 2}{15 \choose 5}}{25 \choose 7}$$However, I am not sure the intuition behind this. I understand that you are choosing 2 red coloured pens from the 10, however, I am not sure why this is. Aren't all the pens the same colour, so why are we treating them as all different? This goes same for the other binomial coefficient chosen.

I also don't get the multiplying and dividing of the binomial coefficients. Also, is there a permutation equivalent for the answer? Thanks.

The argument goes that there are $${25 \choose 7}$$ ways of choosing $$7$$ pens from $$25$$, leading to the denominator. Of these, there are $${10 \choose 2}$$ ways of choosing $$2$$ red pens from $$10$$ and $${15 \choose 5}$$ ways of choosing $$5$$ blue pens from $$15$$, which you multiply together to get the number of ways of getting exactly $$2$$ red pens and $$5$$ blue pens for the numerator. That ignores order.
Alternatively you could choose, in order, the $$7$$ pens from $$25$$ in $$25\cdot 24\cdot 23\cdot 22\cdot 21\cdot 20\cdot 19$$ ways for the denominator. For the numerator, if you chose the $$2$$ red pens and then the $$5$$ blue pens in order there would be $$10\cdot 9\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11$$ ways; but there are $${7 \choose 2}$$ possible orders of $$2$$ red and $$5$$ blue, so you need to multiply by this. It gives the same probability, as both the numerator and the denominator are $$7!$$ times what they were in the earlier calculation, reflecting the number of ways of ordering $$7$$ pens.
• Okay, so for your first sentence: why do we consider each pen as different? Shouldn't we treat all the red pens the same and all the blue pens the same? Therefore, there are technically not ${25 \choose 7}$ ways for choosing 7 pens? – user12055579 Dec 14 '19 at 3:40
• @user12055579 If you try to implement this, you have $25$ physically distinct pens even if many of them look the same. If the question was "what is the probability the first pen drawn is red?" the answer would be $\frac{10}{25} = 0.4$ not $\frac12=0.5$ – Henry Dec 14 '19 at 10:49