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The number of permutations of the letters of the word INDEPENDENT taken 5 at a time are ?

In this type of problem , we have to consider particular cases how a 5 lettered word may be formed. Example , it may be formed by taking all different letters , 3 similar letters and 2 different letters , etc.

while doing a particular case where a word contains 2 similar lettters of one type , 2 similar letters of another type and one different letters , I calculated the number of ways to be $\binom{3}{1}\binom{2}{1}\frac{5!}{2!2!}$ This is because , the letter N is repeated thrice , D is repeated twice and E is repeated thrice in INDEPENDENT. So we can choose the first pair of similar letters in $\binom{3}{1}$ ways , the second pair in $\binom{2}{1}$ ways and the last letter in $\binom{4}{1}$ ways. The word can then be arranged in $\frac{5!}{3!2!}$ ways. But the correct answer to this particular case is $\binom{3}{2}\binom{4}{1}\frac{5!}{3!2!}$ What is wrong in my approach ?

Aditi
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  • I don't understand where $5$ comes into the picture. – lulu Oct 30 '17 at 14:00
  • Sorry, I have edited the question – Aditi Oct 30 '17 at 14:04
  • I see the edit, but it still isn't clear. Do you mean "how many ordered $5-$tuples can be chosen from the letters in $INDEPENDENT$? " or do you mean something else? – lulu Oct 30 '17 at 14:04
  • @lulu I’m sorry I’m not sure what a tuple is. But the 5 letter word that is formed may contain the same letter more than once – Aditi Oct 30 '17 at 14:08
  • You speak about choosing $5$ at a time. That suggests that you are making multiple words, but I think you only intend to make one five letter word. – lulu Oct 30 '17 at 14:10
  • In any case, I can not follow your calculation. As you say, there are three letters that occur at least twice, and seven altogether. But the solo letter need not be chosen from one of the unrepeated letters. I could have the word $NNDDE$ for instance. Not sure how you account for that. – lulu Oct 30 '17 at 14:12
  • @lulu yes I did account for that in another case , but that’s not the particular case that I am talking about. – Aditi Oct 30 '17 at 14:15
  • But I am not sure I am getting the rules right. My count for this case would be: I need to choose two of the three multiples, that's $\binom 32$. Then I need to choose one of the unchosen letters, that's $\binom 51$. Then I need to place the alphabetically first double in my five slots, that;s $\binom 52$. Then I need to place the second double, that's $\binom 32$. But this is not the same as the answer you say is correct so either I am making a blunder (always possible) or I have the rules wrong. – lulu Oct 30 '17 at 14:17
  • I followed the rules you stated. "$2$ similar letters of one type, $2$ similar letters of a different type, and one different letter.". – lulu Oct 30 '17 at 14:18
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    Ah, I see the problem. I can't count. There are only six possible letters. Fine. So then there are $\binom 41$ ways to choose the singleton. Making my answer $\binom 32 \times \binom 41 \times \binom 52\times \binom 32$. Still not sure I have the rules right. – lulu Oct 30 '17 at 14:21
  • See 6-letter permutations in MISSISSIPPI. – r.e.s. Oct 30 '17 at 14:23
  • Right, I had spelled INDEPENDENT incorrectly, so I had $7$ letters. My prior comment contains the corrected version of my count (which still doesn't match the official answer). – lulu Oct 30 '17 at 14:23
  • Okay , let’s wait till someone answers the question. Thanks for the help anyways :) – Aditi Oct 30 '17 at 14:23
  • @lulu's count of $\binom{3}{2}\binom{4}{1}\binom{5}{2}\binom{3}{2}$ is correct. – N. F. Taussig Oct 30 '17 at 14:47
  • In choosing one pair of similar letters and then choosing another, you count each selection twice. For instance, if you first choose D as your repeated letter and then choose E as your other repeated letter, you have the same letters available as you would obtain by first choosing E as your repeated letter and then choosing D as your other repeated letter. – N. F. Taussig Oct 30 '17 at 14:49

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