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Hint: In this question it is given that there are four different letters and four addressed envelopes, so we have to find that in how many ways can the letter be put in the envelopes so at least one letter goes into the correct envelope. So since it is given that at least one of it goes to the correct address i.e, there is a possibility that either 2 letters or 3 letters or 4 letters goes to the correct addressed envelopes. So we have to find the ways for each of the cases.

So find the way we need to know the combination formula, that is, if we want to select r number of quantity from n number of quantity then we can select in $${}^{n}C_{r}$$,

Where, $${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$$ and

$$n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 2\cdot 1$$

Case-1:(1 letter goes to the correct addressed envelope),

So 1 letter can be put in the right envelope from the available 4 envelopes or we can say that among the 4 envelops 1 correct addressed envelope for a letter, can be chosen in $${}^{4}C_{1}$$ ways.

Where,

$${}^{4}C_{1}$$

$$=\dfrac{4!}{1!\cdot \left( 4-1\right) !}$$

$$=\dfrac{4!}{3!} =\dfrac{4\times 3!}{3!} =4$$ ways

Case-2:(2 letters goes to the correct envelopes)

So similarly we can say that 2 letters can be put in the correct envelopes from the available 4 envelopes in $${}^{4}C_{2}$$ ways

Where,

$${}^{4}C_{2}$$

$$=\dfrac{4!}{2!\cdot \left( 4-2\right) !} $$

$$=\dfrac{4!}{2!\cdot 2!} =\dfrac{4\times 3\times 2!}{2\times 1\times 2!} =6$$ ways.

Case-3:(3 letters goes to the correct envelopes)

So we can say that 3 letters can be put in the correct envelopes from the available 4 envelopes in $${}^{4}C_{3}$$ ways.

Where,

$${}^{4}C_{3}$$

$$=\dfrac{4!}{3!\cdot \left( 4-3\right) !} $$

$$=\dfrac{4!}{3!\cdot 1!} =\dfrac{4\times 3!}{3!} =4$$ ways.

Case-4:(3 letters goes to the correct envelopes)

In this case also we can say that 4 letters can be put in the correct envelopes from the available 4 envelopes in $${}^{4}C_{4}$$ ways = 1 ways,

[Since we know that $${}^{n}C_{n}$$ = 1]

Now since each of the cases are independent upon each other then by fundamental principle of addition we can say that the total number of ways that at least one letter goes into the correct envelope = (4+6+4+1) = 15 ways.

Hence the correct option is option A.

Note: In the solution part we have used a principle, which is called fundamental principle of addition or the rule of sum, so in combinatorics, the rule of sum or addition principle is a basic counting principle. Stated simply, it is the idea that if we have A ways of doing something and B ways of doing another thing and we can not do both at the same time(independent actions), then there are A + B ways to choose one of the actions.

So find the way we need to know the combination formula, that is, if we want to select r number of quantity from n number of quantity then we can select in $${}^{n}C_{r}$$,

Where, $${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$$ and

$$n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 2\cdot 1$$

__Complete step-by-step solution:__Case-1:(1 letter goes to the correct addressed envelope),

So 1 letter can be put in the right envelope from the available 4 envelopes or we can say that among the 4 envelops 1 correct addressed envelope for a letter, can be chosen in $${}^{4}C_{1}$$ ways.

Where,

$${}^{4}C_{1}$$

$$=\dfrac{4!}{1!\cdot \left( 4-1\right) !}$$

$$=\dfrac{4!}{3!} =\dfrac{4\times 3!}{3!} =4$$ ways

Case-2:(2 letters goes to the correct envelopes)

So similarly we can say that 2 letters can be put in the correct envelopes from the available 4 envelopes in $${}^{4}C_{2}$$ ways

Where,

$${}^{4}C_{2}$$

$$=\dfrac{4!}{2!\cdot \left( 4-2\right) !} $$

$$=\dfrac{4!}{2!\cdot 2!} =\dfrac{4\times 3\times 2!}{2\times 1\times 2!} =6$$ ways.

Case-3:(3 letters goes to the correct envelopes)

So we can say that 3 letters can be put in the correct envelopes from the available 4 envelopes in $${}^{4}C_{3}$$ ways.

Where,

$${}^{4}C_{3}$$

$$=\dfrac{4!}{3!\cdot \left( 4-3\right) !} $$

$$=\dfrac{4!}{3!\cdot 1!} =\dfrac{4\times 3!}{3!} =4$$ ways.

Case-4:(3 letters goes to the correct envelopes)

In this case also we can say that 4 letters can be put in the correct envelopes from the available 4 envelopes in $${}^{4}C_{4}$$ ways = 1 ways,

[Since we know that $${}^{n}C_{n}$$ = 1]

Now since each of the cases are independent upon each other then by fundamental principle of addition we can say that the total number of ways that at least one letter goes into the correct envelope = (4+6+4+1) = 15 ways.

Hence the correct option is option A.

Note: In the solution part we have used a principle, which is called fundamental principle of addition or the rule of sum, so in combinatorics, the rule of sum or addition principle is a basic counting principle. Stated simply, it is the idea that if we have A ways of doing something and B ways of doing another thing and we can not do both at the same time(independent actions), then there are A + B ways to choose one of the actions.

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