# how many 5 letter words can be formed from letters of INDEPENDENT

Nitya Vootla
37 Points
9 years ago
If you are looking at it from a mathematical point of view~ I’m guessing:
11C5 (selecting 5 letters from given 11)
6C5 (if you do not wish the letters to be repeated)

Y RAJYALAKSHMI
45 Points
9 years ago
But we need to find the number of 5 letter words – not selection
yaswanth
36 Points
9 years ago
11 LETTERS.3N’S 3E’S 2D’S 1P 1I 1T.
NO.OF 5 LETTER WORDS;
CASE 1;OUT OF 5 3ARE ALIKE AND 2 ARE DIFFERENT
2C1.5C2.5!/3!=400
CASE 2;OUT OF 5 3 ARE ALIKE OF ONE KIND AND 2 ARE ALIKE OF OTHER KIND
2C1.2C1.5!/3!.2!=40
CASE 3;OUT OF 5 2 ARE ALIKE OF ONE KIND AND 2 ARE ALIKE OF OTHER KIND AND ONE DIFFERENT LETTER
3C2.4C1.5!/2!.2!=360
CASE 4;OUT OF 5 2 ARE ALIKE 3 ARE DIFFERENT
3C1.5C3.5!/2!=1800
CASE 5;ALL 5 ARE DIFFERENT
6!=720
SO,TOTAL NO.OF 5 LETTER WORDS=400+40+360+1800+720=3320
kartheeek
24 Points
9 years ago
11c5 if repeated and 6c5 if not repeated
Dikshita
11 Points
7 years ago
Wow! What a difficult problem!There are 2 D's, 3 E's, 1 I, 3 N's, 1 P, and 1 TI hope you really meant combinations, and not permutations. For I am notconsidering taking the letters in any order. For example, "E,E,N,N,N", "E,N,E,N,N", "N,N,N,E,E", and "E,N,N,N,E" are all counted as the same combination, even though they would be differentpermutations. Let me know in the thank-you note if you meant permutationsinstead, and I'll re-work the problem.There are 5 basic cases of choosing 5 letters. I will use the last fiveletters of the alphabet, V,W,X,Y and Z as "placeholders" for the letters of acombination. Here are the 5 types of combinations:1. V,V,W,W,W 1 pair, both of the same letters, and 1 triplet, all of the same letter. 2. V,V,W,W,X 2 pairs of same letters and 1 non-matching letter.3. V,V,W,X,Y 1 pair of the same letter and 3 non-matching letters. 4. V,V,V,X,Y 1 triplet, all of the same letter and 2 non-matching letters.5. V,W,X,Y,Z 5 non-matching letters.Case 1. V,V,W,W,WW is more restrictive here than V so we choose it first.We can choose the letter for W in 2 ways, either E or NWe can choose the letter for V in 2 ways, D or whichever one of {E,N} we didn'tchoose in the preceding sentence.That's 2x2 or 4 combinations for case 1.Case 2. V,V,W,W,XWe can choose the 2 letters for the 2 pairs from {D,E,N} in C(3,2) = 3 waysWe can choose the letter for X in any of the remaining 4 ways.That's 3x4 or 12 combinations for case 2.Case 3. V,V,W,X,YWe can choose the letter for the pair 3 ways, D,E, or NWe can choose the letters W,X,Y from any of the 5 remaining ways, in C(5,3) or10 waysThat 3x10 or 30 combinations for case 3.Case 4. V,V,V,X,YWe can choose the letter for V in 2 ways, E or N.We can choose X,Y from any of the remaining 5 letters, in C(5,3) or 10 ways.That's 2x10 or 20 combinations for case 4.Case 5. V,W,X,Y,ZThat's 6 letters taken 5 at a time, or C(6,5) = 6 waysAnswer: 4+12+30+20+6 = 72
Pratik Anil Nimje
13 Points
5 years ago
in the word independent
Repeated letters NNN,DD,EEE.
therefore to find no. Of selection we take different cases.
Case1.when all 5 letters are different
No. Of selection =6C5
Case2.2alike&3different
No. Of selection = 3C1*5C3
Case 3. 3alike&2different
No. Of selection =2C1*5C2
Case4. 3alike & 2alike of different type
No. Of selection =2C1*2C1
Case5 . 2 alike 2like 1different
No. Of selection =3C2*4C1
Therefore total no. Of selection =6C5 + 3C1*5C3 + 2C1*5C2+2C1*2C1+3C2*4C1=
=6+30+20+4+12
=72
natasha
14 Points
5 years ago
this method of solving is good but requires good attention over the different cases possible. here is another method which is simple and also give the right answer
In the word INDEPENDENT there
I=1 , N=3 , D=2 , E=3 , P=1 , T=1
To get five letter words find the coefficient of x^5 in
5![x^0+x^1][x^0+x^1+x^2+x^3][x^0+x^1+x^2][x^0+x^1+x^2+x^3][x^0+x^1][x^0+x^1}
3 years ago
Dear student,

In the word independent
Repeated letters NNN,DD,EEE.
therefore to find no. Of selection we take different cases.
Case1.when all 5 letters are different
No. Of selection =6C5
Case2.2alike&3different
No. Of selection = 3C1*5C3
Case 3. 3alike&2different
No. Of selection =2C1*5C2
Case4. 3alike & 2alike of different type
No. Of selection =2C1*2C1
Case5 . 2 alike 2like 1different
No. Of selection =3C2*4C1
Therefore total no. Of selection =6C5 + 3C1*5C3 + 2C1*5C2+2C1*2C1+3C2*4C1=
=6+30+20+4+12
=72

Thanks and regards,
Kushagra
RISHIKA
550 Points
3 years ago
dear friend