How many toys can be chosen? There are 3 red, 5 blue, 2 yellow and 4 green toys in the box. In how many different ways can 6 toys be chosen if one of them should be blue and the other one - yellow?
I came up with a solution but i am not sure if it is right.
2 toys out of 6 should be of a specific colour. Then, only 4 toys can be of any other colour. 
so, the number of different ways $= C^4_{12} = \frac{12!}{4!8!}=495$

Given the clarifications in comments – the toys are distinct, order doesn't matter, exactly one is blue and exactly one is yellow – we have


*

*$5$ ways to choose the blue toy and $2$ ways to choose the yellow one

*$\binom74=35$ ways to choose the other four toys


Thus there are actually $5×2×35=350$ valid selections.
