IIT JEE , AIEEE Forum - Mathematics , Algebra

akanksha_2606

i am stuck with this question from permutation.plz reply as soon as possible.

Q..find the number of 4 letter words that can be formed using the letters of the word RAMANA?

Am i supposed to take the 3 A's as one unit?or take individually the no. of words

formed when the 1st letter is R,M,N,A etc?

edison

Dear Akanksha

Remember following rules

1) The number of permutations of 'n' different thing takng 'r' at a time without repetition is

ⁿP_r = n!/(n-r)!

So if u go by this approach then in the question there are 4 different letters ( as A is repeated thrice) so ways in which word (containing four letters ) can be formed is

⁴P₄ = 4!/(4-4)! = 4!, what i have suggested in previous response to this question.

2) However if repetition is allowed then the problem can be solved as

There are following cases in which 4 letter words can be formed.

a) 1A, 3 different letters other than A

so for this 3 different letters can be selected from remaining three letters in one way only,

so the ways in which this can be arranged is 4! ....(1)

b) 2A, 2 different letters other than A

Now, if 2 different letters are arranged with 2A's then the permutation is 4!/2!

But the 2 different letters other than A can be selected from 3 letters (R,M & N only) in ³C_²= 3 ways. Taking care of this selection the number of words that can be formed is (4!/2!)*3 .....(2)

c) 3A, 1 different letter other than A

if one different letter is R than the words formed using 3A and R is 4!/3!

But the selection of this different letter from remaining three (R,M,N) can be made in three different ways so total number of words that can be formed is given by (4!/3!)*3 ..... (3)

Thus adding all above possibilities the ways in which 4 letter words can be formed is given by

4! + (4!/2!)*3 + (4!/3!)*3 (ANSWER)

akanksha_2606

THE ANSWER IS GIVEN AS 4!+3.4!/3! +3.4!/2!.SINCE A IS REPEATED THRICE REPETITION IS ALLOWED.

akanksha_2606

sir,

The answer given by you doesn't match with that given in the book.

The answer in the book is given as 4!+3*4!/3!+3*4!/2!.

Nothing is specified about repetition of other letters in the word.' a ' itself is repeated thrice in the word RAMANA.

plz.. reply soon.

puneet

So let us try to solve this problem

Now see we have 4 distinct alphabets in RAMANA.

These are R , A , M , N

And A is repeated 3 times.

So there are 3 cases

Case 1 : The 4 letter word has 1 A

So other 3 letter have to be R, M , N ..

And therefore there are 4! ways .. rite !!

Case 2 : The 4 letter word has 2 A

Now u can choose the rest 2 alphabets from R,M,N

SO total ways = ³C₂* 4! / 2!

³C₂corresponds to choosing 2 alphabets out of 3 and we divide by 2! because there are 2 A's

Case 3 : The 4 letter word has 3 A

Now u can choose the rest 1 alphabet from R,M,N

SO total ways = ³C₁ * 4! / 3!

³C₁corresponds to choosing 2 alphabets out of 3 and we divide by 3! because there are 3 A's .

So total ways are 4! + ³C₂* 4! / 2! + ³C₁ * 4! / 3!

= 4!+3*4!/3!+3*4!/2!.

So now ur answer is matched ... happy ??

cheers

edison

Hope the solution is clear to u by following the approach suggested by Puneet and myself

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