Given word PARASRITJE.How four letter words can be formed by combining them if the last letter is E.
My think is 210 and my friend's think is 8^3
What do you think?!

Supposing the last letter is E, then all you need to count is the number of combinations of A,A,P,R,R,S,I,T,J for three-letter words?

In that case... there are two types of words; abc and aab types (with repetition)

You have A,P,R,S,T,I,J, a total of 7 different letters, so, for "abc" combinations, "a" can be any of the 7, "b" can be any of the 6 remaining and "c" can be any of the last 5. That makes 7 * 6 * 5 = 210 combinations.

A and R are repeated twice, so in aab combinations the "a" can be either A or R (2) and the b can be any of the 6 remaining letters. However, aab can be arranged into: aab, aba and baa, which are three permutations. Therefore we have 2 * 6 * 3 = 36 aab combinations.

So, in total there should be... 210 + 36 = 246 combinations!

it is given that E is the last letter so last place is fixed.then only 3 places are remaining.
and there is only one E in the given word so no E is participating in further arrangement as last place is filled with E.
so i got the answer as 246

@Anas, Off the top of my head, I don't know the answer. I mainly chimed in because I noticed that everyone was solving a different question than form what was stated, and hence gave them a "vote down". Why don't you post how you would approach this problem?

@Harshit, note that the question is about four letter words, while all of your code only displays 3 letter words. Also, your numerical answer of 246 agrees with Estebans (first comment all the way down, which also deals with the 3 letter case).

Yes, but do not forget that there are 2 A's and R's
thus, a word can be AARE for example, withouth repetition, but then we must divide per 2, because A(1)A(2)RE is the same word as A(2)A(1)RE
What do you think, Calvin?

If repetition is allowed, then answer is definitely 8(8)(8) i.e. 512. If repetition is not allowed, then answer is 7(6)(5) i.e. 210.
Explanation can be given for both the first and second answer:-
If repetition is allowed then we can still pick E as a word in the first three places, so still we have eight words to pick in first three places, then The number of configurations is definitely 8(8)(8) i.e. 8(8)(8)=512 words.This verifies the first answer.
But if the repetition is not allowed we have just 7 choices for first letter, 6 choices for the second and 5 choices for the third. Thus by the product rule the answer can be 7(6)(5)=210. This verifies the second answer.
Hope u got it!

Note that his questions specifies for four letter words. Why are the cases only done for 3 letters? You should answer the question as posted.

Aridon, from your suggested answers, I would infer that you mean 3 letter words. Please clarify this point (and also the question raised about allowing for repeats).

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TopNewestSupposing the last letter is E, then all you need to count is the number of combinations of A,A,P,R,R,S,I,T,J for three-letter words?

In that case... there are two types of words; abc and aab types (with repetition)

You have A,P,R,S,T,I,J, a total of 7 different letters, so, for "abc" combinations, "a" can be any of the 7, "b" can be any of the 6 remaining and "c" can be any of the last 5. That makes 7 * 6 * 5 = 210 combinations.

A and R are repeated twice, so in aab combinations the "a" can be either A or R (2) and the b can be any of the 6 remaining letters. However, aab can be arranged into: aab, aba and baa, which are three permutations. Therefore we have 2 * 6 * 3 = 36 aab combinations.

So, in total there should be... 210 + 36 = 246 combinations!

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yes, both of us are right. 210 is wrong, 8^3 is wrong.

It is Esteban G, 16, Spain who is correct. The answer is 246

I'm pretty sure, because I brute forced in python. Here is the code , with all the outputs

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it is given that E is the last letter so last place is fixed.then only 3 places are remaining. and there is only one E in the given word so no E is participating in further arrangement as last place is filled with E. so i got the answer as 246

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Yup, I get 246.

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@Anas, Off the top of my head, I don't know the answer. I mainly chimed in because I noticed that everyone was solving a different question than form what was stated, and hence gave them a "vote down". Why don't you post how you would approach this problem?

@Harshit, note that the question is about four letter words, while all of your code only displays 3 letter words. Also, your numerical answer of 246 agrees with Estebans (first comment all the way down, which also deals with the 3 letter case).

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Anas, I'm not Calvin sir, but i sure do agree with you this is why i think both 210 and 8^3 are wrong.

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Yes, but do not forget that there are 2 A's and R's thus, a word can be AARE for example, withouth repetition, but then we must divide per 2, because A(1)A(2)RE is the same word as A(2)A(1)RE What do you think, Calvin?

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If repetition is allowed, then answer is definitely 8(8)(8) i.e. 512. If repetition is not allowed, then answer is 7(6)(5) i.e. 210. Explanation can be given for both the first and second answer:- If repetition is allowed then we can still pick E as a word in the first three places, so still we have eight words to pick in first three places, then The number of configurations is definitely 8(8)(8) i.e. 8(8)(8)=512 words.This verifies the first answer. But if the repetition is not allowed we have just 7 choices for first letter, 6 choices for the second and 5 choices for the third. Thus by the product rule the answer can be 7(6)(5)=210. This verifies the second answer. Hope u got it!

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Note that his questions specifies for four letter words. Why are the cases only done for 3 letters? You should answer the question as posted.

Aridon, from your suggested answers, I would infer that you mean 3 letter words. Please clarify this point (and also the question raised about allowing for repeats).

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The second answer can be correct too, if repetitions are allowed. (For example, EEEE.) I think the question is not properly specified.

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