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No. of 3 letters words can be made from the word INEQUALITIES without repetition.


So i did this, i separated unrepeted letters from repeted ones, which looked like..
repeatation= {I,I,I,E,E} & without repetition= {Q,L,N,U,A,T,S}

SO, no of words made from 7 letters without repetition = ⁷P4 = 210

& Number of 3 letter words without repetition that can be made from repetition = 5! / 3! 2! = 10

Total= 210+10= 220, So answer seems right but i am not sure of approach! Any guidance will help very much.
EDIT1:
Answer is right i.e 220.
EDIT2: Without repetation of Letters.
220 claimed by my misprinted book is wrong answer. "504" is the right answer.
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if followup question would be to find with repetitions of letters than 553 is the answer.

kunal Ch.
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2 Answers2

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Addendum added to respond to OP's comment.

I agree with coffeemath's comments. Your $\displaystyle \frac{7!}{4!}$ computation represents 3 letter words that exclude any "I" or "E" letters. I'm unsure what your $\displaystyle\binom{5}{2}$ computation is then supposed to represent.

Personally, I would break the situation into mutually exclusive cases, letting $T_k$ denote the computation for Case $k$.

$\underline{\text{Case 1:} ~3~ \text{I's are used.}}$
This can only occur with the specific word III.
$T_1 = 1.$

$\underline{\text{Case 2: A double letter was used.}}$
There are $2$ ways of choosing whether to use $2$ I's or $2$ E's.
Assume that $2$ I's are used. Then, there are 8 ways of selecting one letter from
{ENQUALTS}.
Once the off letter is chosen, there are 3 positions that the off letter may be placed in.
Therefore,
$T_2 = 2 \times 8 \times 3 = 48.$

$\underline{\text{Case 3: Only 1 of each type of letter was used.} }$
Then, there are $\displaystyle\binom{9}{3} = 84$ ways of selecting $3$ distinct letters from
{EINQUALTS}.
Once these three letters are selected, they can be permuted in $3! = 6$ ways.
Therefore, $T_3 = 84 \times 6 = 504.$

Final computation:

$$T_1 + T_2 + T_3 = 1 + 48 + 504 = 553.$$

Addendum
Based on the OP's comment, Cases 1 and 2 above are disallowed. This implies that, since only Case 3 is allowed, the final computation should be $T_3 = 504$.

Again, I am at a loss as to how to reverse engineer an interpretation that justifies the computation of $(220)$ as the final answer.

user2661923
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  • The motive behind my calculation is to reach the answer which is 220! I understood each part of your calculation but am not sure as your answer differs. --True blue anil 's answer also differs from your's as well as books, but i found your ones more understandable. – kunal Ch. May 12 '21 at 12:12
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    @kunalCh. I am at a loss to reverse-engineer an interpretation of the problem, for which the right answer is $(220)$. Where my interpretation differs from true blue anil's interpretation is that I think that the intent of the problem composer is that the letters are to be treated like "scrabble tiles", where if you have two "E" tiles, you can use both of the tiles. Why else present a word with $3$ I's and $2$ E's? Just to create a poorly worded trick question? – user2661923 May 12 '21 at 12:17
  • This problem already exists, in my book. Pardon me on behalf of my book. @user2661923 – kunal Ch. May 12 '21 at 12:24
  • I realized just now, your computation is fine with repetition of letters. We are intrested in without repetition of letters hence words like III or 2 letters of same kind cases won't arise. – kunal Ch. May 12 '21 at 12:40
  • I agree that the "scrabble tiles" interpretation is better. "Without repetition" could just mean without reselecting a particular already-selected letter. – Joffan May 12 '21 at 13:14
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Well, if the exact wording is "without repetitions"
I would interpret it as no letter being used more than once in the word, which makes it quite easy.

There are $9$ distinct letters, $I\,N\,E\,Q\,U\,A\,L\,T\,S$

thus number of number of $3$ letter words = $P^9_3 = 9\cdot8\cdot7$

true blue anil
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    Reasonable interpretation, but begs the question: if this is the intended interpretation, why choose a string of letters that includes "III" and "EE". Further, the OP, upon seeing the very first comment posted, did not immediately respond by saying that "EEQ" was disallowed. On the other hand, I agree that the problem's wording is ambiguous. – user2661923 May 12 '21 at 11:13
  • @user2661923: Yes, the wording is ambiguous, that's why I asked OP whether he had reproduced the exact wording, and he replied affirmatively. – true blue anil May 12 '21 at 11:43
  • The exact wording is: The no of 3 letters word that can be made from the word INEQUALITIES if repetitions of letters is NOT allowed – kunal Ch. May 12 '21 at 12:35
  • --user2661923 i corrected that error, no words like NQN,EEQ are not allowed. – kunal Ch. May 12 '21 at 12:37