How many 4-element DNA sequences containg exactly three of four bases A,T,C,G?
Hello, my answer is c(4,3)*c(3,1)*(4!/2!) because
by saying c(4,3), I choose three of four bases, e.g. ATC_
by saying c(3,1), I choose extra one of chosen three bases, e.g. ATCT
by saying 4!/2!, I permutated them, i.e., there is 4!/2! possible permutations of ATCT.
So, I found the answer as 144, but there is a solution on the internet like this:
The objective is to find the 4 element DNA sequence that contain exactly three of the four bases A, C, G, and T.
Three of the four bases, but not with base A.
There are 3 ways for first base as it cannot be A.
There are 3 ways for second base as it cannot be A.
There are 2 ways for third base as it cannot be A and nor first or second base.
There are 1 way for fourth base as it cannot be A and nor first or second or third base.
By product rule, 3.3.2.1 =18.
Exactly three of four bases.
As there are 4 possible bases that are excluded and there are 18 DNA sequences without corresponding base but with three of the four bases.
By product rule, 4.18 = 72.
Therefore, there are 72 possible way for 4 DNA bases.
I wonder which answer is correct 144 or 72? and if my solution with the answer 144 is correct, then why is the solution on the internet with answer 72 incorrect? or if my solution is incorrect, then why?
There are four ways to exclude a base. There are three choices for which of the remaining bases shows up twice. There are $4!/2$ ways to arrange $4$ items when $2$ of them are identical. Given that, there are $4 \cdot 3 \cdot \frac{24}{2} = 144$ ways.
Check: let's do the cases removing G. There should be $36$ of them. Let's say the A is doubled. There should be $12$ of those:
With AA together: AACT
AATC
CAAT
TAAC
CTAA
TCAA
With AA separated:
CATA
TACA
ACTA
ATCA
ACAT
ATAC