how many four letter word using word ineffective
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In the word ineffective
There are 2 -i
1 -n
3-e
2-f
1-c
1-t
1-v
So to make a four letter word
Possibilities are
1) 3 alike one different ---- means three e and one any other letter out of I,n ,c ,t ,v,f
So now choose 3 e by 1c1 and then one other letter by 6 c1 = 1c1*6c1
Now to permute these by 4!/3!
Since 3 are alike so answer of 1 st case = 1c1*6c1*4!/3! = 24
2) two alike and two other alike
There are two e ,I,f are possible so to choose three out of two by 3c2 then permute it by 4!/2!*2! Since two alike and two other alike answer = 3c2*4!/2!*2! = 18
3) 2 alike two different
Now to choose two alike by 3c1 then two different by 6c2
And permute it by 4!/2!
Thus = 6c2 *4!/2! = 180
4) all four are different
So choose four from 7 by 7c4 = 35
And permute it by 4!= 24
Thus = 35*24 = 840
Now add all four cases = 24+18+180+840 = 1062
This is required result
There are 2 -i
1 -n
3-e
2-f
1-c
1-t
1-v
So to make a four letter word
Possibilities are
1) 3 alike one different ---- means three e and one any other letter out of I,n ,c ,t ,v,f
So now choose 3 e by 1c1 and then one other letter by 6 c1 = 1c1*6c1
Now to permute these by 4!/3!
Since 3 are alike so answer of 1 st case = 1c1*6c1*4!/3! = 24
2) two alike and two other alike
There are two e ,I,f are possible so to choose three out of two by 3c2 then permute it by 4!/2!*2! Since two alike and two other alike answer = 3c2*4!/2!*2! = 18
3) 2 alike two different
Now to choose two alike by 3c1 then two different by 6c2
And permute it by 4!/2!
Thus = 6c2 *4!/2! = 180
4) all four are different
So choose four from 7 by 7c4 = 35
And permute it by 4!= 24
Thus = 35*24 = 840
Now add all four cases = 24+18+180+840 = 1062
This is required result
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