In how many different ways can $3$ red, $4$ yellow and $2$ blue bulbs be arranged in a row?

I think we are to assume bulbs of the same colour are indistinguishable. So we are counting the "words" of length $9$ that have $3$ R, $4$ Y, and $2$ B.

The places for the R's can be chosen in $\binom{9}{3}$ ways. For each of these ways, the places for the Y's can be chosen in $\binom{6}{4}$ ways. Multiply and simplify.

About the books, imagine that that the math books are placed in a box, labelled M. Then we have $9$ objects, the English books and the M. These can be arranged on the shelf in $9!$ ways. For each of these ways, the math books can be taken out of the box and arranged in $4!$ ways, for a total of $9!4!$.

For the probability, divide as you did by $12!$.