Step-by-step explanation:
the six letters a, b, c, d, e, f have to be arranged without repetition. but the sequence matters, of course.
so, for the first letter we have 6 possibilities.
after picking the first letter, we have now 5 possibilities for the second letter. then 4 for the 3rd letter, 3 for the 4th letter, 2 for the 5th letter and then one remaining option for the 6th.
that gives us
6! = 6×5×4×3×2×1 = 720 different ways of arranging the 6 letters without repetition.
can it be that you misread 720 for 120 ?
or did you mix up the questions ?
or did you hold back some other information about restrictions ?
for example 120 would be the right answer about how many different groups of 3 letters you can create out of the 6 without repetition, and the sequence matters (e.g. a, b, c is different to c, b, a).
then these are permutations without repetitions.
P(6, 3) = 6! / (6-3)! = 6×5×4×3×2 / 3×2 = 6×5×4 = 120.
120 would be also the correct answer to the problem to have 1 fixed starting letter, and the other 5 can be freely arranged. then we have
1×5×4×3×2×1 = 120 different ways of arranging the letters without repetitions.
so, what is the actual problem ?
the one you described has the right answer 720.
NOT 120.
