How many possible ways are there to pick three cards with the number one in them out of 20 cards labeled 1-20?

Question

Theres 20 cards and each of them is labeled with the numbers 1 to 20 . In how many ways can you pick 3 cards ehich include 3 1s? Ex.(11,2,12) ; (19,18,10). Also (11,2,12) , (12,2,11) and (11,2,11) are 3 different ways so order matters. My proffesor said that there is 7560 different ways that you can pick them , but im getting 810 ways. Im splitting it in two different cases: Case 1:One of the cards is 11 , so one of the cards contains 1 and the other doesnt , so i get 1×10×9(since only one way to get 11 , 10 different ways for cards with 1 and 9 without 1). 90 is the result. Case 2:None of the numbers is 11 so all 3 cards contain 1 . We have 10 possible ways for the first one , 9 for the second and 8 for the third , so i get 10×9×8=720. Then i sum both of the results and i get 720+90=810. Im not sure what my mistake here is . Thank you in advance.

Answer 1

I can't reverse engineer the Professor's answer.

Your work has some mistakes:

There is 1 number with 2 1's: 11.

There are 10 #'s with a single 1 : $\{1,10,12,13,14,15,16,17,18,19\}.$

There are 9 #'s with no 1 : $\{2,3,4,5,6,7,8,9,20\}.$

When selecting the # 11 : there are $10 \times 9 = 90$ combinations.

When not selecting the # 11 : there are $\binom{10}{3} = 120$ combinations.

I get final answer of $(90 + 120) \times 3! = 210 \times 3! = 1260.$

Answer 2

There are $10 \times 9 \times 6=540$ three card permutations where one card is $11$, one card has one $1$ and one card had no $1$s.

There are $10 \times 9 \times 8 = 720$ three card permutations where each card has one $1$.

This makes a total of $1260$ permutations with three $1$s.

Your professor seems to have multiplied the answer by another factor of $6$.

Answer 3

As there are $20*19*18 = 6840 < 7560$ ways total to pick $3$ cards the professor can't be right.

I have to assuming you are allowed replacement.

In which case there are $20^3 = 8000$ ways to to pick three cards. But there are $9^3 = 729$ to pick cards with no ones. So there only $8000-729 = 7271$ ways to pick three cards with any number of ones.

So there is no way your professor is right and you reported the question as intended.