Binomial and direct probability calculation

Question

i have this problem:

a careless bookseller mixture 4 broken books with 16 normal ones from one specific book. Four friend go to his library to buy this book. a) Calculate the probability of 3 of the 4 friends take the broken books and one get the normal one. b) What is the probability of after the friends go to these library it remains with the same amount of broken books?

If i try to solve this problem by bernoulli, we have:

a) if $X$ = number of broken books

$x ~ B(n, p) = {n \choose k} \cdot p^k \cdot (1 - p)^{n-k}$

$x ~ B (20, 1/5)$

if $k$ = number of success of the Bernoulli, which is get a broken book

$B(20, 1/5) = {20 \choose 3} \cdot (1/3)^3 \cdot (4/5)^{17} = 0,20536$

And if i calculates the probability by hand, i got:

$B$ = broken book

$N$ = normal book

i want $BBBN$

so:

$P = \frac{4}{20} \cdot \frac{3}{19} \cdot \frac{2}{18} \cdot \frac{16}{17} \cdot \frac{4!}{3!} = 0,013209$

Does anyone knows why im getting this difference? what im doing wrong?

As the letter B is the same as letter A, i just post the question for letter A

Answer 1

Comment continued: Rushing to a meeting; no time for a proper exposition. Here are two versions of R code that solve it. Maybe you can 'decode' it.

dhyper(3, 4, 16, 4)
## 0.01320949
choose(4,3)*choose(16,1)/choose(20,4)
## 0.01320949

---

Added later, after Acceptance, for completeness.

Total number of ways in which 4 friends can each select a book from among 20 is ${20 \choose 4} = \frac{20}{4!\, 16!} = 4845.$ Total number of ways in which 3 friends can each select a damaged book from among 4 is ${4 \choose 3} = 4.$ Total number of ways in which 1 friend can select a normal book from among 16 is ${16 \choose 1} = 16.$

Probability exactly 3 friends select damaged book and 1 selects a normal book is $\frac{64}{4845} = 0.01320949.$

Let $X$ be the number of friends who happen to select damaged books. It has the hypeergeometric distribution tabled below.

 x = 0:4;  pdf = dhyper(x, 4, 16, 4)
 cbind(x, pdf)
     x          pdf
  ## 0 0.3756449948
  ## 1 0.4623323013
  ## 2 0.1486068111
  ## 3 0.0132094943  <-- Answer
  ## 4 0.0002063983