# Number of combinations with limited repetition

 P: 771 For the first atom,there is M choices...for the second,again M choices...for the third,again M choices...and so on! So there is always M possible choices and all that we should do is to multiply the number of choices for each of the atoms which becomes $M^n$. But because the atoms are identical,we should decrease this amount by dividing it by $n!$.