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

The **Answer to the Question**

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**Here's the Solution to this Question**

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.