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Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6. This one is clearly insufficient: we don't know how many letters are repeated in the word or even how many letters are there. Not sufficient.

(2) There are 5 letters in the word. We don't know how many letters are repeated in the word. Not sufficient.

(1)+(2) We can deduce that the last three letters of the word are all different (hence their arrangement of 3! = 6) but we still don't know whether they repeat any of the first two letters. For example, if the word is goose, then the number of arrangements of its letters would be 5!/2! but if the word is close, then the number of arrangements of its letters would be 5!. Not sufficient.

Answer: E.

Try NEW Combinations PS question. _________________

Re: Certain word is written on a paper. What is the number of ar [#permalink]

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26 Jul 2014, 16:56

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I believe the correct answer is E, solely for the fact that either statement doesn't disclose whether or not there are duplicate letters in the word.

Statement 1) if two letters were omitted there would be six possible combinations. This statement makes me assume that there are three unique letters that can combined in 3x2x1 ways, however there could be more than three letters where some are duplicates and form a total of six possible combinations. We also don't know about the first two letters.

Statement 2) there's five letters, clearly insufficient. Don't know if there are duplicates. Could be 5! Or 5!/2! Ect.

Statements together : I still don't know the first two letters. I know three of the five letters are unique from our first statement however the first two could be the same or different. Total combinations could be 5! Or 5!/2!

Re: Certain word is written on a paper. What is the number of ar [#permalink]

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26 Jul 2014, 19:16

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i think ans would be E bcoz 1. given that if first two letters are deleted, no of arrangements are 6 ie 3! based on this we know that totally we have 5 letters but we don't know anything about first two letters that means first two letters would be alike or unlike.(5!/2! or 5!) 2. no information regarding how many letters alike or unlike

Even though we use both options, we don't have any information about first two letters.

Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6 (2) There are 5 letters in the word

Kudos for a correct solution.

Key to answering this question is identifying how many Unique Letters there are assuming no other restrictions are noted.

Unfortunately, neither statements actually address whether all letters are unique or not.

Statement 1: This says that the word has at least 3 unique letters. It does not say anything whether the other letters are also unique Statement 2: This only gives us the number of letters. We do not know whether any of the letters have duplicates.

Combined: Neither statements address the question whether the first 2 letters are duplicates or not.

Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6 (2) There are 5 letters in the word

Kudos for a correct solution.

1) Insufficient. case i: the word has 8 letters and in last 6, 5 are repeating. so after removing first 2 letters, arrangement for last 6 = 6!/5! = 6 case ii: the word has 5 letters and in last 3 none is repeating. so after removing first 2, arrangement for last 3 = 3! = 6

2) Insufficient. 5 letters could mean 5!, 5!/2!, 5!/3!,...

Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6 (2) There are 5 letters in the word

Kudos for a correct solution.

Statement 1: even if we equate 8 to a permutation nPr and get the value ...we still don't know wether the first two letters are identical or not. INSUFFICIENT statement 2: knowing that there are 5 letter is again not enough to get a permutation. all the letters might be identical or all can be different. INSUFFICIENT.

taking both together: still not sufficient coz no information about the indentical letters.

Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6. This one is clearly insufficient: we don't know how many letters are repeated in the word or even how many letters are there. Not sufficient.

(2) There are 5 letters in the word. We don't know how many letters are repeated in the word. Not sufficient.

(1)+(2) We can deduce that the last three letters of the word are all different (hence their arrangement of 3! = 6) but we still don't know whether they repeat any of the first two letters. For example, if the word is goose, then the number of arrangements of its letters would be 5!/2! but if the word is close, then the number of arrangements of its letters would be 5!. Not sufficient.

Answer: E.

Kudos points given to correct solutions.

Try NEW Combinations PS question. _________________

Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6 (2) There are 5 letters in the word

Kudos for a correct solution.

1) Insufficient. case i: the word has 8 letters and in last 6, 5 are repeating. so after removing first 2 letters, arrangement for last 6 = 6!/5! = 6 case ii: the word has 5 letters and in last 3 none is repeating. so after removing first 2, arrangement for last 3 = 3! = 6

2) Insufficient. 5 letters could mean 5!, 5!/2!, 5!/3!,...

(1) + (2) Insufficient arrangements = 5! or 5!/2!

so E.

These are not the only possible cases:

For abcde it would be 5!. For aacde it would be 5!/2!. For cccde it would be 5!/3!. For cdcde it would be 5!/(2!2!).
_________________

Re: Certain word is written on a paper. What is the number of ar [#permalink]

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30 Mar 2015, 00:49

1) after omitting the first 2, remaining arrangements are 6. This is possible in many ways : eg : if there are 3 letters left (3! = 6), also if there are letters left out of which 2 are repeated (4! / 2!*2! = 6). Thus more than one sol => NS. 2) Total = 5 letters. We dont know if any of those is repeated or not. Thus NS.

1 and 2 => Again whether letters are repeated or not, is not known. Thus the first 2 omitted letters can be same or different or same as one of the other letters. Thus total arrangements cant be calculated. Ans E.