# 9 letter of the alphabet

1. Suppose we want to arrange the first 9 letters of the
alphabet: A, B, C, D, E, F, G, H, I (every letter is used exactly
once). In how many ways can this be done

(a) if there are no restrictions?

(b) If the word BAG must appear in the arrangement. (For
example, FDCBAGEHI, BAGCDEFHI, etc)

(c) if all of the vowels appear consecutively and all of the
consonants appear consecutively? (For example, AIEBDGHF, HGDBCFEAI,
etc.)

(d) if the vowels must all appear consecutively? (For example,
BCAEIGFHD, HGDBCFEAI, etc.)

e) if at least two of the first three letters in the arrangement
are vowels?

2. Given 10 French books, 20 Spanish books, 8 German books, 15
Russian books and 25 Italian books,

(a) How many books must be selected to guarantee there are 2
books of the same language?

(b) How many books must be chosen do guarantee there are 7 books
of the same language?

(c) How many books must be chosen do guarantee there are 12
books of the same language?

3. Show that if n + 1 numbers are chosen from 1, 2, . . . , 2n,
then two of the numbers must always be consecutive integers.

1.

Total number of ways to arrange 9 different alphabets = 9!

2.

In this consider {BAG} as a single alphabet, than total number
of ways to arrange 7 different alphabetswill be

= 7!

3. Now consider {AEI} as a single alphabet and {BCDFGH} as a
single alphabet

Then total different alphabet = 2

Number of way in which we can arrange two different alphabets =
2!

Now

Number of ways in which we can arrange {AEI} = 3!

Number of ways in which we can arrange {BCDFGH} = 6!

Total number of ways = 2!*3!*6!

D.

Now consider {AEI} as a single alphabet

Then total different alphabet = 7

Number of way in which we can arrange two different alphabets =
7!

Now

Number of ways in which we can arrange {AEI} = 3!

Total number of ways = 7!*3!

E.

Total number of case = Case when all first three are vowels +
when at least two from first three are vowels

1. Case when all first three are vowels

from part C

Total cases = 3!*6!/2!

2. Cases when exact two from first three is vowels

Words will be of three type = CVVCCCCCV +
VVCCCCCCV + VCVCCCCCV

Number of ways will be for CVVCCCCCV

= 6C1*3C2*6!

C1 is number of ways to choose the first Constant (C)

3C2 is number of ways to choose the next Vowels(V)

6! is number of ways to choose the rest of 6 alphabets

Similiarly for other two cases the number of ways will be
same.

Total Number of ways = 3!*6!/2! + 3*6C1*3C2*6! = 3*6!*(1 +
6C1*3C2) = 57*6!