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The number of ways in which 5 letters ca...

The number of ways in which 5 letters can be placed in 5 marked envelopes, so that no letter is in the right encelope, is

A

45

B

44

C

43

D

46

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The correct Answer is:
To solve the problem of finding the number of ways in which 5 letters can be placed in 5 marked envelopes such that no letter is in the correct envelope, we will use the concept of dearrangements. A dearrangement is a permutation of the elements of a set, such that none of the elements appear in their original position. ### Step-by-Step Solution: 1. **Understand the Problem**: We need to find the number of ways to arrange 5 letters in 5 envelopes such that no letter goes into its corresponding envelope. This is a classic example of a dearrangement. 2. **Use the Formula for Dearrangement**: The number of dearrangements \( !n \) of \( n \) items can be calculated using the formula: \[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \] For our case, \( n = 5 \). 3. **Calculate \( 5! \)**: First, we need to calculate \( 5! \): \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] 4. **Calculate the Summation**: We will compute the summation \( \sum_{i=0}^{5} \frac{(-1)^i}{i!} \): - For \( i = 0 \): \( \frac{(-1)^0}{0!} = 1 \) - For \( i = 1 \): \( \frac{(-1)^1}{1!} = -1 \) - For \( i = 2 \): \( \frac{(-1)^2}{2!} = \frac{1}{2} \) - For \( i = 3 \): \( \frac{(-1)^3}{3!} = -\frac{1}{6} \) - For \( i = 4 \): \( \frac{(-1)^4}{4!} = \frac{1}{24} \) - For \( i = 5 \): \( \frac{(-1)^5}{5!} = -\frac{1}{120} \) Now, we sum these values: \[ 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \] To simplify this, we can find a common denominator, which is 120: \[ = \frac{120}{120} - \frac{120}{120} + \frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120} \] \[ = 0 + \frac{60 - 20 + 5 - 1}{120} = \frac{44}{120} \] 5. **Final Calculation**: Now, substituting back into the dearrangement formula: \[ !5 = 5! \cdot \left(\frac{44}{120}\right) = 120 \cdot \frac{44}{120} = 44 \] Thus, the number of ways to arrange 5 letters in 5 envelopes such that no letter is in the correct envelope is **44**.

To solve the problem of finding the number of ways in which 5 letters can be placed in 5 marked envelopes such that no letter is in the correct envelope, we will use the concept of dearrangements. A dearrangement is a permutation of the elements of a set, such that none of the elements appear in their original position. ### Step-by-Step Solution: 1. **Understand the Problem**: We need to find the number of ways to arrange 5 letters in 5 envelopes such that no letter goes into its corresponding envelope. This is a classic example of a dearrangement. 2. **Use the Formula for Dearrangement**: The number of dearrangements \( !n \) of \( n \) items can be calculated using the formula: \[ ...
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OBJECTIVE RD SHARMA ENGLISH-PERMUTATIONS AND COMBINATIONS-Chapter Test
  1. The number of ways in which 5 letters can be placed in 5 marked envelo...

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  2. 7 women and 7 men are to sit round a circulartable such that there is ...

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  3. There are (n+1) white and (n+1) black balls, each set numbered 1ton...

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  4. 12 persons are to be arranged to a round table. If two particular pers...

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  5. The number of committees of 5 persons consisting of at least one femal...

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  6. The number of ways in which a team of eleven players can be selected f...

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  7. In a football championship, 153 matches were played. Every two-team pl...

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  8. How many numbers between 5000 and 10,000 can be formed using the digit...

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  9. If x, y and r are positive integers, then ""^(x)C(r)+""^(x)C(r-1)+""^(...

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  10. In how many ways can 5 red and 4 white balls be drawn from a bag conta...

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  11. All the letters of the word 'EAMCET' are arranged in all possible ways...

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  12. There are 10 lamps in a hall. Each one of them can be switched on i...

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  13. How many 10-digit numbers can be formed by using digits 1 and 2

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  14. The straight lines I(1),I(2),I(3) are parallel and lie in the same pla...

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  15. about to only mathematics

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  16. The number of diagonals that can be drawn by joining the vertices of a...

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  17. The sum of the digits in unit place of all the numbers formed with the...

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  18. In an examinations there are three multiple choice questions and each ...

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  19. There are 10 points in a plane, out of these 6 are collinear. If N is ...

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  20. Ramesh has 6 friends. In how many ways can be invite one or more of th...

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  21. If Pm stands for ^m Pm , then prove that: 1+1. P1+2. P2+3. P3++ndotPn=...

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