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

Y RAJYALAKSHMI, 9 years ago

Grade:12th pass

9 Answers

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

Now add all cases

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}

Kushagra Madhukar

askIITians Faculty 628 Points

3 years ago

Dear student,

Please find the attached solution to your problem below.

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

Now add all cases

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

my answer is 55440

Since there are no constraints related to repetition or other. So, you can have “55440” words of 5 letters from the letters of the word INDEPENDENT.

THANK U