Q. 85.0( 4 Votes )

# Mark the correct alternative in the following:

The number of arrangements of the word “DELHI” in which E precedes I is

A. 30

B. 60

C. 120

D. 59

Answer :

Detailed Solution

Considering {E} and {I} as single object, we have {D}, {EI}, {L}, {H}.

The number of words that can be formed out of the letters of the word “DELHI” in which {E} precedes {I} is = 4!

= 24. (When {E} is just before {I})

However, it is not required to place {E} just before {I}, so, in that case, the solution will be:

__Case-1:__ Let us say, the first letter of the word be {E}, so the word format becomes E××××, where the ‘×’ are to be filled with the other 4 letters, this can be done in 4! = 24 ways.

The discussion can be shown pictorially as:

__Case-2:__ Let us say, the second letter of the word be {E}, so the word format becomes ×E×××, where the first ‘×’ is to be filled with anyone of the 3 letters i.e. {D}, {L}, {H} [not with {I}], this can be done in 3 ways.

And, the other 3 ‘×’ are to be filled with the remaining 3 letters, this can be done in 3! = 6 ways.

So, in this case total number of arrangements

= 18.

The discussion can be shown pictorially as:

__Case-3:__ Let us say, the third letter of the word be {E}, so the word format becomes ××E××, where the first 2 ‘×’ are to be filled with any 2 of 3 letters i.e. {D}, {L}, {H} [not with {I}], this can be done in = 6 ways.

And, the other 2 ‘×’ are to be filled with the remaining 2 letters, this can be done in 2! = 2 ways.

So, in this case total number of arrangements

= 12.

The discussion can be shown pictorially as:

__Case-4:__ Let us say, the third letter of the word be {E}, so the word format becomes ×××E×, where the first 3 ‘×’ are to be filled with 3 letters i.e. {D}, {L}, {H} [not with {I}], this can be done in 3! = 6 ways.

And, the last ‘×’ is to be filled with the remaining {I}, this can be done only in 1 way.

So, in this case total number of arrangements

= 6.

The discussion can be shown pictorially as:

As, the cases are independent, so, the total number of arrangements of the word “DELHI” in which E precedes I is = 24 + 18 + 12 + 6

=__60__.

__ALTERNATIVE METHOD__

We have, 5 objects {D}, {E}, {L}, {H}, {I}.

So, the number of words that can be formed out of the letters of the word “DELHI” is = 5!

= 120.

From, symmetry, we can say, out of these 120 words in 60 words {E} precedes {I} and in other 60 words {I} precedes {E}.

So the required answer = __60__.

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Which of the following are true :

(2 + 3)! = 2! + 3!

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