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Permutations & Combinations

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CAT 2025 Lesson : Permutations & Combinations - Derangement

Derangements of

\bm{r}

objects is when each of these objects are misplaced or wrongly assigned.

Number of Derangements

=

\bm{(r)}

r! \times \left( 1 - \dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + ... \right)

In cases where there is a total of

\bm{n}

objects, and

\bm{r}

of them are misplaced or wrongly assigned, we first select the

r

elements in

^{n}C_{r}

ways and then derange them.

Number of such derangements

=

^{n}C_{r} \times D(r) =

^{n}C_{r} \times r! \times \left( 1 - \dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + ... \right)

It's not too hard to derive the derangements. However, we recommend you to memorise the following.

(1) = 0

(2) = 1

(3) = 2

(4) = 9

(5) = 44

Furthermore, derangements can be derived as follows.
When

n

is odd, D

(n) = (

(n - 1) \times n) - 1

When

n

is even, D

(n) = (

(n - 1) \times n) + 1

Using this we can derive D

(6) = 44 \times 6 + 1 = 265

; D

(7) = 265 \times 7 - 1 = 1854

and so on.

What is the number of ways in which

6

letters can be put into

6

envelopes, such that exactly

2

of the letters get into the correct envelopes?

From

6

letters,

4

need to be selected and put into incorrect envelopes, i.e. deranged.

Number of ways

=

^{6}C_{4} \times

(4) = 15 \times 9 = 135

Answer:

135

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