we know The genetic code is based on the four nucleotides adenine (A), cytosine (C), guanine (G), and thymine (T). These are connected in long strings to form DNA molecule.
with three A, one C, two G and one T how many distinct string with size 7 can be formed?
i try to solve it. i think it's 4^7 or 420 ?
any idea or hint would be highly appreciated.
On
No $4^7$ would be correct (and the value is much larger) if you had lots of each base, but here you have exactly seven bases available. First imagine the seven bases were distinct-how many orders would that be? Now take three of the seven and make them interchangeable. How many orders that were different are now the same? Finally make another pair interchangeable.
On
This is just a multinomial coefficient, i.e. $$ \binom{7}{3, 1, 2, 1} = \frac{7!}{3! \, 1! \, 2! \, 1!} = 420 $$ Explanation is given by numbering the repeat symbols, so you have 7 different symbols and just $7!$; if you erase the numbering you see that the 3 $A$'s, for instance, mean you overcounted by a factor of $3!$. Repeat for each of the other letters, and you get the multinomial coefficient.
You can choose the places where the 3 $A$s will go in $7C3$ ways, then, for each of these arrangements, the $G$s can be put in in any of the 4 places left in $4C2$ ways , then $C$ can be put in the 2 places left in $2$ ways, and then the $T$ is put in the remaining place. The total is the product of these. Note that this result is independent of the order in which you select the aminoacids.
The total is then $$7C3 \times 4C2 \times 2 \times 1= \frac{7!4!(2)(1)}{3!4!2!2!}=420 $$