Answer

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**Hint:**The letters present in ANNIHILATION are A, N, I, H, L, O and T. The number of A in the word is 2, number of N is 3, Number of I is 3 and rest of letters occur only once first we will write the number of possibilities to write 4 as sum of 2 or more numbers and then find the total number of permutations possible.

**Complete step by step solution:**

Let’s note down all possibilities to write 4 as sum of 2 or more integers

1. 1+ 1+ 1+ 1 = 4

2. 1+ 1+ 2 = 4

3. 2+ 2 = 4

4. 1+ 3 = 4

In the first condition we have to choose 4 different letters and arrange them, so the number of possibilities to choose 4 numbers out of 7 and arrange them is equal to ${}^{7}{{C}_{4}}\times 4!=840$

In the second condition we have to choose 2 different and 2 same number, we know that A, N, I occur more than once so we can choose one of them so total possibilities will be ${}^{3}{{C}_{1}}\times {}^{6}{{C}_{2}}\times \dfrac{4!}{2!}=540$

In the third case we have to choose a pair of same number, so we have to choose from A, N, I

So the number of total possibilities will be ${}^{3}{{C}_{2}}\times \dfrac{4!}{2!2!}=18$

In fourth case we have to choose 3 same number, N and I occur thrice so total possibilities will be ${}^{2}{{C}_{1}}\times {}^{6}{{C}_{1}}\times \dfrac{4!}{3!}=48$

So the total number of possibilities will be equal to 840+ 540+ 18+ 48 = 1446.

**Note:**

These types of problems can be solved by taking different cases because we do not know which letter will come how many times and we have to arrange the multiple occurring letters so it can only be solved by taking cases.

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