Find the number of permutations of the first 8 letters of the alphabet for taking 1 letter at a time?

The first 8 letters of the alphabet are:

{A, B, C, D, E, F, G, H}

To find the number of permutations, we can think of 8 positions. In each one of these positions, we can put one of the letters, now we need to find the number of possible options for each one of these positions, the total number of permutations will be equal to the product between all of these options.

Now, taking one letter at a time (to assign at each position)

For the first position, we have 8 options.

For the second position, we have 7 options (because one letter is already taken)

For the third position, we have 6 options (because two letters are already taken, one in the first position and another in the second)

For the fourth position, we have 5 options.

For the fifth position, we have 4 options.

For the sixth position, we have 3 options.

For the seventh position, we have 2 options.

For the last position, we have 1 option.

Then the total number of permutations is:

P = 8*7*6*5*4*3*2*1 = 40,320

step-by-step explanation:

homozygous. one is homozygous, while two is heterozygous.