Name: Principle of Inclusion and Exclusion
Uploaded: 2019-07-19
Duration: 18 min 3 s
Description: The inclusion-exclusion principle is an important combinatorial way to compute the size of a set or the probability of complex events. It relates the sizes o...

Question

Find the number of words those can be formed by using all letters of the word 'DAUGHTER'. If
(i) Vowels occurs in first and last place.
(ii) Start with letter

G

and end with letter

H

.
(iii) Letters

G, H, T

always occurs together.
(iv) No two letters of

G, H, T

are consecutive
(v) No vowel occurs together
(vi) Vowels always occupy even place.
(vii) Order of vowels remains same.
(viii) Relative order of vowels and consonants remains same.
(ix) Number of words are possible by selecting

2

vowels and

3

consonants.

Accepted Answer

Given word is ''DAUGHTER :Total words $= 8$

Vowels are $' A, U, E'$

Consonants are $' D, G, H, T, R'$

(i) Vowels occur in first and last place

For the first letter, there are $3$ possibilities

So, after filing the first letter, the last letter has $2$ possibilities

Filing the other $6$ places, we can do that in $6!$ ways because all letters are different.

So, the number of ways = $3 \times 2 \times 6! = 4320$ ways

(ii) Start with letter $G$ and end with letter $H$

Remaining places can be filled in $6!$ $= 720$ ways.

(iii) Letters $G, H, T$ occurs together always

Consider $GHT$ as one object, then the number of words $= 5 + 1 = 6$

Number of ways of arranging them $= 6!$ ways

We can re-arrange $G, H, T$ among themselves in $3!$ ways.

So, the number of ways $= 3! \times 6! = 4320$ ways.

(iv) No $2$ letters $G, H, T$ are consecutive

Case ( $1$ ): All $3$ are consecutive

Number of ways $= 4320$ ways

Case ( $2$ ): Only $G, H$ are consecutive

Consider $GH$ as one object, then the number of objects $= 6 + 1 = 7$

Numbers of arranging them $= 7! \times 2!$ ways

But in these, we also get the cases in which $G, H, T$ come together, so we need to subtract $4 \times 6!$

The number of arrangements in which the only $G, H$ are consecutive $= 7! \times 2! - 4 \times 6!$

Total number of ways in which at least $2$ of $G, H, T$ are consecutive $= 4320 + 3 (7! \times 2! - 4 \times 6!) = 25920$

Number of ways of arranging such that no two ways in which at least $2$ of $G, H, T$ are consecutive

$=$ Number of ways $- 25920$

$= 8! - 25920$

$= 14400$ ways

(v) No vowel occur together

$3$ vowels- $A, U, E$ .

The same argument as above follows for this.

So, number of ways $= 14400$

(vi) Vowels occupy even space

There are $4$ even places and $3$ vowels, so we can arrange vowels in ${}^{4}P_{3}$ ways.

The remaining $5$ letters can be arranged in $5!$ ways.

So, number of ways $= {}^{4}P_{3} \times 5! = 2880$ ways,

(vii) Order of vowels remains same

The total number of arrangements $= 8!$ ways

But in these, we cannot change the order of vowels.

Suppose the required ways is $' n'$ . So, multiplied by $3!$ ways with $' n'$ must be given $8!$

So, $3! \times n = 8!$

$∴ n = 6720$ ways.

(viii) Order of vowels and consonants remains same.

Then the same way, there are $5$ consonants and $3$ vowels

So, $5! \times 3! \times n = 8!$

$∴ n = 56$ ways.

(ix) We need to select two vowels which can be done in ${}^{3}C_{2}$ ways. Selecting $3$ consonants can be done in ${}^{5}C_{3}$ ways.

Rearranging can be done $5!$ ways

So, number of words that are possible by selecting $2$ vowels and $3$ consonants is ${}^{5}C_{3} \times {}^{3}C_{2} \times 5!$

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