1. ## 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 $\displaystyle \: = \frac {8!}{2!}\: =\: 20160.$

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

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

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

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

What next?

Thanks.

ILoveMaths07.

2. Hello, ILoveMaths07!

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

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

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

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

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

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

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

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

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

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