# Permutations

• Sep 20th 2008, 08:22 AM
ILoveMaths07
Permutations
Calculate the number of permutations of the word CALENDAR in which C and A are together but N and D are not.

This is what I did:

No. of possible permutations $\: = \frac {8!}{2!}\: =\: 20160.$

No. of permutations in which C and A are together $\: =\: 5! * \frac {3!}{2!}\: =\: 360.$

No. of permutations in which C and A are not together $= 20160 - 360 = 19800.$

No. of permutations in which N and D are together $\: = \frac {6!}{2!} * 2! = 720.$

No. of permutations in which N and D are not together $= 20160 - 720 = 19440.$

Thanks.

ILoveMaths07.
• Sep 20th 2008, 12:11 PM
Soroban
Hello, ILoveMaths07!

Quote:

Calculate the number of permutations of the word CALENDAR
in which $C$ and $A$ are together, but $N$ and $D$ are not.

First, find the number of permuations in which $C$ and $A$ are together.

Duct-tape $C$ and $A$ together.
They could be: $\boxed{CA}$ or $\boxed{AC}$ . . . 2 ways.

We have 7 "letters" to arrange: . $\{A,D,E,L,N,R, \boxed{CA}\, \}$
. . They can be arranged in: . ${\color{blue}7!}$ ways.

Hence, there are: . $2 \times 7! \:=\:10,080$ ways to have $C$ and $A$ together.

How many have $C$ and $A$ together and $N$ and $D$ together?

Duct-tape $C$ and $A$ together. There are 2 ways: . $\boxed{CA}$ or $\boxed{AC}$

Duct-tape $N$ and $D$ together. There are 2 ways: . $\boxed{ND}$ or $\boxed{DN}$

We have six "letters" to arrange: . $\{A,E,L,R,\boxed{CA},\,\boxed{ND}\, \}$
. . They can be arranged in: ${\color{blue}6!}$ ways.

Hence, there are: . $2\times 2 \times 6! \:=\:2880$ ways to have $C,A$ and $N,D$ together.

Therefore, there are: . $10,080 - 2,880 \;=\;\boxed{7,200}$ ways
. . in which $C$ and $A$ are together, but $N$ and $D$ are not.