How many different words can be formed by using all the letters of the word ALLAHABAD if both l do not come together?

Free

10 Questions 10 Marks 10 Mins

Concept:

Let there be n things of which p1 are alike of one kind, p2 are alike of another kind, p3 are alike of 3rd kind, ..…, pr are alike of rth kind such that p1 + p2 + ….+ pr = n. Then the permutations of n objects is \(\rm \frac{n!}{(p_1!)\times (p_2!)\times ....\times (p_r!)} \)

Calculation:

The word ALLAHABAD contains 9 letters, in which A occur 4 times, L occurs twice and the rest of the letters occur only once.

Number of different words formed by the word ALLAHABAD using all the letters

\(\rm =\frac{9!}{4!\times2!}=\frac{9\times8\times7\times6\times5\times4!}{4!\times2}\)

\(=\rm \frac{72\times7\times30}{2}\)

= 7560

Now, let us take both L together and consider (LL) as 1 letter

Then we will have to arrange 8 letters, in which A occurs 4 times and the rest of the letters occur only once

So, the number of words having both L together \(=\rm \frac{8!}{4!}=\frac{8\times7\times6\times5\times4!}{4!}\)

= 1680

Hence, the number of words with both L not occurring together

= 7560 - 1680

= 5880

Hence, option (3) is correct.

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.3 Crore+ Students

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Skip for now

How many different words can be formed by using the letters of the word ‘ALLAHABAD?

(a) In how many of these do the vowels occupy even positions.

(b) In how many of these, the two L’s do not come together?

Number of letters in word 'ALLAHABAD' = (A → 4, L → 2, H → 1, B → 1, D → 1) = 9

Number of arrangements =

(a) There are only four A's as vowels.

They can occupy even places (2, 4, 6, 8) in

ways

∴ Number of ways in which vowels occupying even places = 1We are left with 5 places and letters (L → 2, H → 1, B → 1, D → 1).

Number of permutations =

Hence, total number of arrangements in which A's occupy even places = 1 x 60 = 60.(b) We first find the number of arrangements in which two L's are not together: Number of arrangements in which two L's are together

Hence, the number of arrangements in which the two L's are not together = (Total arrangements) - (the number of arrangements in which the two L's are together)

= 7560 - 1680 = 5880.

HI, thanks for this wonderful stuff. I have a concern with last part of this question where you multiplied 1680 by saying both L can be arranged in two ways by themselves. Is it correct if both is the same word? Kindly reply if you really have started this forum for the visitors, not for the sake of "Link Building for search engines"

Best Regards