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Jodi M.
Algebra
8 months, 1 week ago
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Jii How many ways we can arrange the letters of word independent
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Video Transcript
in this problem, we have been given the word independent and we have to determine the number of ways in which we can arrange all the letters of this word. Independent. So first we observed that there are in total 11 letters So we can arrange these 11 distinct letters in 11 factorial ways. But here we see that some of the letters they are repeated. Like we observed that I hear is just appearing once And here that's repeated three times. And de here that's repeated two times and e it's repeated three times in this word. And thus we conclude that the number of possible arrangements that's equal to 11 factorial over Three Factorial Times two Factorial Times three Factorial. And when we simplify this we can get the results coming out to be double five double four double zero. So these are the different ways by which we can arrange the letters of the word independent. So that's the required answer.
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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
Find the total number of ways of selecting five letters from the letters of the word INDEPENDENT.
Solution
INDEPENDENT
11 letters (3N, 3E, 2D), I, P, T = 6 types.
We have to form 5 letters - word
I. All different =
6C5.5!=(6).5!=720
II. 2 alike, 3 diff. =
3C1.5C3.5!2!
=(3.10).60=1800.
III. 3 alike, 2 diff. = 2C1.5C2.5!3!
=(2.10).20=400.
IV. 2 alike, 2 alike, 1 diff. =
3C2.4C1.5!2!2!
=(3.4).30=360.
V. 3 alike, 2 alike =
2C1.2C1.5!3!2!
=(2.2).10=40
Total selections =
6+30+20+12+4=72
Total words = 3320.