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.
 A: 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.    
Both approaches are correct and give the same probability.  
