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

Question

$1)$ How many different ways can $3$ red, $4$ yellow and $2$ blue bulbs be arranged in a row?

Do I just say $3! 4! 2! = 288$ ?

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$2)$ On a shelf there are $4$ different math books and $8$ different English books.

a. If the books are to be arranged so that the math books are together, how many ways can this be done?

b. What is the probability that all the math books will be together?

For part $a)$, I put $8! 4! = 967680$ and part $b)$, $\dfrac{8!4!}{ 12!}$

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I'm not too sure if i did these right and I would appreciate some help, thanks.

Answer 1

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 $12!$.

Answer 2

$1)$ Suppose the red, yellow, and blue bulbs are identical, then the total number of ways to arrange them in a row is: $\dfrac{9!}{3!4!2!}=...$

$2)$ There are $9$ ways for the math books to be together,and for each of these ways, you have $4!8!$, thus the total number of ways for the math books to be together is: $9\times 4!\times 8! = ...$

$3)$ The probability that all the math books will be together is: $\dfrac{9\times 4!\times 8!}{12!}= ...$