Answer
Verified
40.8k+ views
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
$$n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 2\cdot 1$$
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.
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.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
Other Pages
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
A food packet is dropped from a helicopter rising up class 11 physics JEE_Main
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Write dimension of force density velocity work pre class 11 physics JEE_Main
Amongst LiCl RbCl BeCl2 and MgCl2the compounds with class 11 chemistry JEE_Main