De arrangements

De arrangements

If n distinct objects are arranged in a row, then the number of ways in which they can be de arranged so that none of them occupies its original place is.

\(n!\left[ 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-….+{{(-1)}^{n}}\frac{1}{n!} \right]\).

And it is denoted by D(n)

If r(0 ≤ r ≤ n) objects occupy the place assigned to them i.e., their original places and none of the remaining (n – r) objects occupies its original places, then the number of such ways is

D (n – r) = ⁿCᵣ.D(n – r)

= ⁿCᵣ. (n – r)!\(n!\left[ 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-….+{{(-1)}^{n-r}}\frac{1}{(n-r)!} \right]\).


Ajay writes letters his five friends and addresses the corresponding. The number of ways can the letters be placed in the envelops so that at least two of them are in the wrong envelops are.

Solution: Required number of ways =\(=\sum\limits_{r=2}^{5}{^{5}{{C}_{5-r}}D(r)}\).

\(=\sum\limits_{r=2}^{5}{\frac{5!}{r!(5-r)!}.r!\left[ 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+….+\frac{{{(-1)}^{n}}}{r!} \right]}\).

\(=\sum\limits_{r=2}^{5}{\frac{5!}{r!(5-r)!}.r!\left[\frac{1}{2!}-\frac{1}{3!}+….+\frac{{{(-1)}^{n}}}{r!} \right]}\).

\(=\frac{5!}{3!}\left( \frac{1}{2!} \right)+\frac{5!}{2!}\left( \frac{1}{2!}-\frac{1}{3!} \right)+\frac{5!}{1!}\left( \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} \right)+\frac{5!}{0!}\left( \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!} \right)\).

= 10 + 20 + (60-20+5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 199