Kenneth Rosen, exercise 6.1, Qs - 42 (d)

Question

How many 4-element DNA sequences contain exactly three of the four bases A, T, C, and G?

Solution given:

There are four ways to choose which letter is to occur twice and three ways to decide which of the other letters to leave out, so there are 4 · 3 = 12 choices of the letters for the sequence. There are 4 positions the first (alphabetically) of the single-use letters can occupy, and then 3 positions for the second single-use letter, a total of 4 · 3 = 12 different sequences once we have determined the letters and their frequencies. Therefore the answer is 12 · 12 = 144.

 

My approach:

No. of ways in which I can choose 3 out of 4 bases = 4C3.

Now, no. of ways in which I can arrange the 3 chosen bases such that one base occurs twice = 4C3 * ( 4! / 2! ) = 4 * 12 = 48.

 

Please do let me know if my approach is wrong. It would be of great help if you can show what combinations my approach is not including but the given solution includes.

Comments

As you said, “No. of ways in which I can choose 3 out of 4 bases” i.e. $\binom{4}{3}$: $\{A,T,C\},\{A,T,G\},\{A,C,G\},\{T,C,G\}$

Now, you have to make 4-element DNA sequences from the elements of the above 4 sets by repeating any of the 3 elements from a set. So there are $3$ ways to repeat an element because there are $3$ elements in the set.

For example, for the set $\{A,T,C\},$ you can repeat either $A,T$ or $C$ to make a sequence of 4 elements.

So, 4 ways to make sets of 3 elements and then 3 ways to repeat an element from each set and then $\dfrac{4!}{2!}$ to make 4-element sequence from 2 same elements out of 4 elements.

Hence total possible 4-element DNA sequences are: $4 \times 3 \times \dfrac{4!}{2!}$