How many distinct permutations are there of the letters of the word "MARRIAGE"? a.30240 b.40320 c.20160 d.10080

We have two letters which repeat in the anagram of the word MARRIAGE.

The number of distinct permutations is $\frac{8!}{2!2!}=8*7*6*5*3*2=10080$ .

We can receive this result with other way. When we arrange letters to 8 places we can begin from two letters A. There are 7+6+5+4+3+2+1=28 ways to do this. When we arranged A, then we can arrange two letters R to 6 residual places. This number is 5+4+3+2+1=15. When we arranged R, we have 4 residual places to arrange letters M, I, G, E. The number of this arrangements is 4!=4*3*2=24. Total number or permutations is

28*15*24=10080.