four-digit numbers ?

Hi, i got some problem to understand this question. Need some help please..

How many four-digits numbers contain only the digits 1 and 2 and each of them at least once?

(A) 10 (B) 12 (C) 14 (D) 15 (E) 16

:?:

You could work by cases. If the number has exactly one 1, how many numbers can you form with that? You know you have four digits, four "slots" to fill. How many ways can you place the 1? Etc, etc.

If you get stuck, please reply showing all of your work and reasoning. Thank you!

Eliz.

Hello, nelsonling!

How many four-digits numbers contain only the digits 1 and 2 and each of them at least once?

. . \(\displaystyle (A)\;10\;\;\;(B)\;12\;\;\;(C)\;14\;\;\;(D)\;15\;\;\;(E)\;16\)

Click to expand...

Since it seems that there are at most 16 of them, you could trying listing them.

We are forming a four-digit number.
. . For the first (leftmost) digit, we have two choices: use a "1" or a "2".
. . For the second digit, we have two choices.
. . For the third digit, we have two choices.
. . For the last digit, we have two choices.

Hence, there are: \(\displaystyle \,2\,\times\,2\,\times\,2\,\times\,2\:=\:16\) choices for the four-digit number.

But this includes "1111" and "2222",
. . and the problem asks for numbers that have both a "1" and a "2" (at least).

Therefore, there are: \(\displaystyle \:16\,-\,2\;=\;14\) such numbers.

Great thanks Soroban