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Ask iit jee aieee pet cbse icse state board experts Expert Question: permutation and combination
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ruhi (603)

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hi
plzzzzz explain how to do this question
"how many 4 letter words can be formed using the letters of the word "INEFFECTIVE"? ''
plzzzzzzzz i need urgent reply
ruhi
    
puneet1622 (38)

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The word INEFFECTIVE has following letters
 
                    E,E,E ; F,F ; I,I ; N,C,T,V
 
So the number of 4 letter words will be the coefficient of x4 in the expression
 
              4! ( 1 +  x/1 + x2/2 + x3/6)(1 +  x/1 + x2/2)2(1 + x)4
where the first term corresponds to how many times E would be taken. The x terms are divided by the factorial of the number of times they are taken because the alike letters will result in less permutations.
The second term corresponds to the number of times the F and I are taken and the last term corresponds to the number of times N,C,T,V are taken.
 
So find the coefficient of x4 to obtain the correct answer.
 
This is a general method and can be applied to any word.
 
Cheers
 

There is only one .. one best way !!!!!!!!!
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puneet (3521)

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Great work done this a absolutely correct approach !!!!!!!!!!
 
I hope ruhi this helps u !!

Puneet Agrawal
IIT Delhi
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ruhi (603)

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hi
is there any oyher method to solve this question???
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ruhi (603)

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or plz explain this method in a more simple manner
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puneet (3521)

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Well yes there is one more method where u actually take case and see that whether we have taken all different letters or some letters are alike and so on.
 
However the method described above is simple as u do not need to worry about the cases. The answer will automatically come. DO you understand the approach that has been used above ??

Puneet Agrawal
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ruhi (603)

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sir i ve not used this method before so i could not get it. can u plz explain it 2 me????
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puneet (3521)

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okay let me explain ... first of all do not use that word SIR .. just a kid like u ... may made it earlier to IIT thats it . ok
 
Now about the approach .. see the letter E can appear in a four letter word for one , two or three times.. OKay ?? Now if the letter E appears 3 times then we need to divide by factorial(3) since otherwise we are counting redundant case.
So note that the x^3 has been divided by factorial(3) which is 6.
Similarily for other letters. Try to understand that x^n denotes that the letter appears n times in the word. Now is it clear ???
 
 

Puneet Agrawal
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ruhi (603)

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sorry but still not getting
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ruhi (603)

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ok ok now i got it.
thanx
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puneet1622 (38)

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hehe cannot understand this ...

There is only one .. one best way !!!!!!!!!
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puneet (3521)

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Ooooooops hmm how do I explain ?? hehe ... try to take a simple example let us suppose the word in LAMA and we need to make 2 letter words from that. OKAY ??
 
So let us see ... by using the previous approach
 
We will find coefficeint of x2 in the expression
    2! (1 + x)2(1 + x + x2/2)
   
    So we find coefficeint of x2 in ( 1 + x2 + 2.x).(1 + x + x2/2)
 
    The terms corresponding to x2 are 2! ( 1 + 2.1 + 1/2)x2
    which is 7.
 
     Now see the normal way the two letter words can be 
AA, AL, AM, LA, LM, MA, ML
 
which are 7 in number so the approach works okay??
 
The expression will change if u make 5 letter words .. in this case u will replace 4! by 5! and find coefficiet of x^5.
 
okay ... cheers

Puneet Agrawal
IIT Delhi
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ruhi (603)

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now i got it
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puneet (3521)

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Thank god u got it ........

Puneet Agrawal
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gopi_mith (106)

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puneet sir has used an excellent method which is actually an application of the multinomial theorem

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