Suppose that in a deck of 20 cards, each card has one of the number 1, 2, 3, 4, 5 with four cards of each number. Ten cards are chosen from the deck at random. If the cards are drawn with replacement, what is the probability that each of the numbers 1, 2, 3, 4, 5 will appear exactly twice?
This is a problem from my class 12 workbook, the problem probably has something to do with considering the events independents and conditional probability since these are all the topics I have been taught in class and expected to use.
Alternative approach:
$\underline{\text{Overview}}$
Initially, I (reluctantly) did not post an answer, because the posted question is of low quality, with respect to this article on MathSE protocol.
I would have been reluctant because the OP had left a comment where he obliquely referred to cases. I consider that my solution, which does involve cases, to be beyond what you would require of a Math student new to the topic.
Then, I read JMoravitz' answer which is much more elegant than mine. However, in my opinion, in order to understand JMoravitz' answer, you have to intuit the concept in his first paragraph. That is, you would have to understand that you can presume that you are working with a $5$ card deck.
Although I can grasp the concept after reading it, my comprehension is nowhere near strong enough to have originated the presumption. I think that this presumption is also beyond what you might reasonably expect from a Math student new to the topic.
So, once JMoravitz posted, I decided to bend the rules and demonstrate how a Math student new to the topic might reasonably attack the problem.
$\underline{\text{Helper function} ~f(k)}$
Assume that with each of the $5$ ranks, you have $(4)$ distinguishable cards, one of each of the following suits: Spades; Hearts; Diamonds; Clubs.
For $k \in \{0,1,2,3,4,5\}$ let $f(k)$ denote the number of satisfying distributions of $(10)$ cards, where in $k$ of the ranks, the card of a specific suit has been selected twice. This implies that it is being assumed that in $(5-k)$ of the ranks, two cards of different suits are being selected.
So, you have $(6)$ mutually exclusive cases, depending on how many ranks had the same card selected twice.
Therefore, the overall probability is
$$\frac{\sum_{k=0}^5 f(k)}{20^{(10)}}.$$
$\underline{\text{Computation of} ~f(k)}$
As a compromise, I will manually compute $~f(0), f(5),~$ and $~f(3)$. I will then use these computations to derive a pattern, and use this pattern to provide a generic formula for $f(k)$.
To compute $f(0)$, you are assuming that $(0)$ of the ranks had the same suit selected twice. Then, there are $~\displaystyle \left[\binom{4}{2}\right]^5~$ ways of selecting $10$ different cards, and then $(10)!$ ways of permuting these $(10)$ cards.
Therefore,
$$f(0) = \left[\binom{4}{2}\right]^5 \times (10)!.$$
To compute $f(5)$, you are assuming that $(5)$ of the ranks had the same suit selected twice. Then, there are $~\displaystyle \left[\binom{4}{1}\right]^5~$ ways of selecting $5$ different cards, where each card is selected twice. Then, as discussed in the answer of JMoravitz, you have a multinomial distribution.
Therefore,
$$f(5) = \left[\binom{4}{1}\right]^5 \times \binom{10}{2} \times \binom{8}{2} \times \binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}$$
$$= \left[\binom{4}{1}\right]^5 \times (10)! \times \frac{1}{2^5}.$$
To compute $f(3)$, you are assuming that $(3)$ of the ranks had the same suit selected twice. First of all, there are $~\displaystyle \binom{5}{3}~$ ways of determining which of the $(5)$ ranks will have the same card selected twice. Then, there are $~\displaystyle \left[\binom{4}{1}\right]^3 \times \left[\binom{4}{2}\right]^2~$ ways of selecting the $7$ different cards. That is, in $3$ of the ranks, your are selecting $1$ card each, where each card is selected twice, and then in $2$ of the ranks you are selecting $2$ different cards. Then, you have a partial multinomial distribution.
Therefore,
$$f(3) = \binom{5}{3} \times \left[\binom{4}{1}\right]^3 \times \left[\binom{4}{2}\right]^2 \times \binom{10}{2} \times \binom{8}{2} \times \binom{6}{2} \times (4!)$$
$$= \binom{5}{3} \times \left[\binom{4}{1}\right]^3 \times \left[\binom{4}{2}\right]^2 \times (10)! \times \frac{1}{2^3}.$$
There is now a clear pattern. For $~k \in \{0,1,2,3,4,5\},$
$$f(k) = \binom{5}{k} \times \left[\binom{4}{1}\right]^k \times \left[\binom{4}{2}\right]^{(5-k)} \times (10)! \times \frac{1}{2^k}. \tag1 $$
$\underline{\text{Overall Computation}}$
The overall probability is
$$\frac{\sum_{k=0}^5 \left\{ ~\binom{5}{k} \times \left[\binom{4}{1}\right]^k \times \left[\binom{4}{2}\right]^{(5-k)} \times (10)! \times \frac{1}{2^k} ~\right\}}{20^{(10)}} $$
$$= \frac{\sum_{k=0}^5 \left\{ ~\binom{5}{k} \times \left[\binom{4}{1}\right]^k \times \left[\binom{4}{2}\right]^{(5-k)} \times (10)! \times \frac{1}{2^k} ~\right\}}{5^{(10)} \times 4^{(10)}}. \tag2 $$
$\underline{\text{Equivalence with the answer of JMoravitz}}$
The answer of JMoravitz is
$$\dfrac{10!}{2^5\cdot 5^{10}}. \tag3 $$
I can reduce the challenge of showing that (2) and (3) above are equivalent by removing the common factor of
$$\dfrac{10!}{5^{10}}.$$
The challenge is then reduced to showing that
$$\dfrac{4^{(10)}}{2^5} = \sum_{k=0}^5 \left\{ ~\binom{5}{k} \times \left[\binom{4}{1}\right]^k \times \left[\binom{4}{2}\right]^{(5-k)} \times \frac{1}{2^k} ~\right\}. \tag4 $$
Since $~\displaystyle \binom{4}{2} = (3 \times 2),~$ the RHS of (4) above may be re-written as
$$\sum_{k=0}^5 \left\{ ~\binom{5}{k} \times \frac{4^k}{2^k} \times 3^{(5-k)} \times 2^{(5-k)} ~\right\} $$
$$= \sum_{k=0}^5 \left\{ ~\binom{5}{k} \times 2^k \times 3^{(5-k)} \times 2^{(5-k)} ~\right\} $$
$$= \sum_{k=0}^5 \left\{ ~\binom{5}{k} \times 2^k \times 3^{(5-k)} \times \frac{2^5}{2^k} ~\right\} $$
$$= 2^5 \times \sum_{k=0}^5 \left\{ ~\binom{5}{k} \times 3^{(5-k)} ~\right\}. $$
The challenge that was presented in (4) above has been simplified to showing that
$$\dfrac{4^{(10)}}{2^5} = 2^5 \times \sum_{k=0}^5 \left\{ ~\binom{5}{k} \times 3^{(5-k)} ~\right\}. \tag5 $$
However, the assertion in (5) above may be immediately seen to be true, by binomial expansion. That is
$$4^5 = (3 + 1)^5 = \sum_{k=0}^5 \left\{ ~\binom{5}{k} \times 3^{(5-k)} ~\right\}.$$