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Total number of ways in which 3 letters can be arranged is 5*4*3 = 60 ways Total number of ways 3 letters can be arranged such that A and B are always together is # of ways A and B can be arranged between themselves (2)* # of ways in which the remaining 3 letters can be picked (3) * # of ways in which the 3 letters and (A&B) can be arranged(2) = 3*2*2 = 12 Therefore, the number of ways in which A and B are never together = 60 - 12 --> 48

However, the OA is 42. What am I missing?

Also, in general (particularly in data sufficiency), can we assume that the elements will always be arranged in a straight line?

We are asked to find the total number of ways in which 3 out of the 5 alphabets A, B, C, D and E can be arranged in a straight line such that A and B never come together.

In order to arrange 3 alphabets, we have to select 3 alphabets from the given 5. This can be done in (5x4)/2 = 10 ways However, out of these 10 selections, we need only those selections wherein A and B are not together. Let's first find the number of selections in which A and B are selected. In this case, out of the 3 alphabets to be selected, A and B are always in. That means, we just need to select 1 more alphabet from the remaining 3 i.e. C, D and E. This can be done in 3 ways. So the total number of selections in which A and B are together = 3 Therefore, of the total 10 selections, only 7 are such that A and B are not selected.

So, the number of ways of selecting 3 alphabets such that A and B are not selected = 7 ways.

Now, we are asked to arrange these 3 alphabets in a straight line. We know that 3 alphabets can be arranged in a straight line in 3! = 3x2 = 6 ways.

Hence, the total number of arrangements which do not include A and B together is 7x6 = 42 ways.

Hope this helps! _________________

MBA Candidate 2015 | Georgetown University McDonough School of Business

Thanks for the reply Sudish. I'm a little confused though as to why you've treated this as a combination rather than as a permutation? _________________

Total number of ways of arranging 3 things out of 5 is 5*4*3 = 60. (5p3) Lets take the cases in which A and B are always selected. So total ways will be (1*1*3) (one for a, one for b and select any one out of C,D, and E). But these things can also be arranged in 3*3! ways = 18. 3! ways to arrange A,B and any 1 other alphabet multiplied by 3 to select either D,E or F.

Thanks for the explanations. I still did not get why the answer is 42 and not 48.

I agree with total no. of permutations = 5*4 * 3 = 60

Considering the cases when A & B are in the final three, we have the following 18 combinations

AB* ABC ABD ABE BA* BAC BAD BAE *AB CAB DAB EAB *BA CBA DBA EBA

A*B ACB ADB AEB B*A BCA BDA BEA

This is also obtained by (1*1*3)*3! = 18 ways

But the question says only that A & B should not be together in a straight line. So the last 6 - A*B and B*A should be also ok with the results. If that is true, then it should be only 12 ways that are not allowed( and not 18). This can also be arrived at by considering in the following way: A and B can be arranged within in 2 ways (AB) can be arranged with 3 other letters DEF in 3 ways And position of AB and D/E/F can be interchanged in 2 ways.

Hence, 2*3*2 = 12 ways.

Final answer should be 60-12 = 48, based on the question. If it said that A&B together should not be in the final three at all( or A& B cannot be together in a circle positioning), then it is 60 - 18 = 42

Please let me know if I am making any mistake or assumption here.

Thanks for the explanations. I still did not get why the answer is 42 and not 48.

I agree with total no. of permutations = 5*4 * 3 = 60

Considering the cases when A & B are in the final three, we have the following 18 combinations

AB* ABC ABD ABE BA* BAC BAD BAE *AB CAB DAB EAB *BA CBA DBA EBA

A*B ACB ADB AEB B*A BCA BDA BEA

This is also obtained by (1*1*3)*3! = 18 ways

But the question says only that A & B should not be together in a straight line. So the last 6 - A*B and B*A should be also ok with the results. If that is true, then it should be only 12 ways that are not allowed( and not 18). This can also be arrived at by considering in the following way: A and B can be arranged within in 2 ways (AB) can be arranged with 3 other letters DEF in 3 ways And position of AB and D/E/F can be interchanged in 2 ways.

Hence, 2*3*2 = 12 ways.

Final answer should be 60-12 = 48, based on the question. If it said that A&B together should not be in the final three at all( or A& B cannot be together in a circle positioning), then it is 60 - 18 = 42

Please let me know if I am making any mistake or assumption here.

I think the question means that A and B are never selected together in a selection. So it should be 18 ways. You'll have to consider A*B and B*A cases also.

Number of ways of getting a 3 letter code out of 5 distinct letters A,B,C,D and E such that A and B never come together :

Three mutually exclusive cases are possible:

Case a: The 3 letter code will contain A but not B Case b: The 3 letter code will contain B but not A Case c: The 3 letter code will contain both A and B but not togather

Case a: The 3 letter code will contain A but not B First letter A, secong any letter other than A or B, third any letter other than A or B and the already chosen letter 1x3x2 = 6 But A could be 1st or 2nd or 3rd. Therefore we have 3 ways of getting this. Therefore 6x3= 18

Case b: The 3 letter code will contain A but not B same as case a. Therfore 18

Case c: The 3 letter code will contain both A and B but not togather First Letter A , 2nd any letter other than A or B, Third Letter B 1x3x1 = 3 But The first letter can be either A or B. Therefore we have 2 cases 3x2=6

Number of ways of getting a 3 letter code out of 5 distinct letters A,B,C,D and E such that A and B never come together :

Three mutually exclusive cases are possible:

Case a: The 3 letter code will contain A but not B Case b: The 3 letter code will contain B but not A Case c: The 3 letter code will contain both A and B but not togather

Case a: The 3 letter code will contain A but not B First letter A, secong any letter other than A or B, third any letter other than A or B and the already chosen letter 1x3x2 = 6 But A could be 1st or 2nd or 3rd. Therefore we have 3 ways of getting this. Therefore 6x3= 18

Case b: The 3 letter code will contain A but not B same as case a. Therfore 18

Case c: The 3 letter code will contain both A and B but not togather First Letter A , 2nd any letter other than A or B, Third Letter B 1x3x1 = 3 But The first letter can be either A or B. Therefore we have 2 cases 3x2=6

Total no of ways = 18 + 18+ 6= 42

There is one more case, depending on how you read it Case [d]. The three letter code will not contain A or B.

that is: C, D, E. Number of permutations = 3! = 6.

But agree with others, the question is not clear enough.

Thanks for the explanations. I still did not get why the answer is 42 and not 48.

I agree with total no. of permutations = 5*4 * 3 = 60

Considering the cases when A & B are in the final three, we have the following 18 combinations

AB* ABC ABD ABE BA* BAC BAD BAE *AB CAB DAB EAB *BA CBA DBA EBA

A*B ACB ADB AEB B*A BCA BDA BEA

This is also obtained by (1*1*3)*3! = 18 ways

But the question says only that A & B should not be together in a straight line. So the last 6 - A*B and B*A should be also ok with the results. If that is true, then it should be only 12 ways that are not allowed( and not 18). This can also be arrived at by considering in the following way: A and B can be arranged within in 2 ways (AB) can be arranged with 3 other letters DEF in 3 ways And position of AB and D/E/F can be interchanged in 2 ways.

Hence, 2*3*2 = 12 ways.

Final answer should be 60-12 = 48, based on the question. If it said that A&B together should not be in the final three at all( or A& B cannot be together in a circle positioning), then it is 60 - 18 = 42

Please let me know if I am making any mistake or assumption here.

You are right. I would mark the answer as 48 and not 42. If anything, the question clearly asks "In how many ways can they be arranged so that they do not come together?" So I would say ACB is fine but ABC is not.

Arrange 3 letters out of 5 in 5*4*3 = 60 ways and arrange 3 letters out of 5 such that A and B are there and together in 3*2!*2 = 12 ways Required number of cases = 60 - 12 = 48 _________________

Re: In how many ways can 3 letters out of 5 distinct letters A [#permalink]
18 Feb 2014, 03:50

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In how many ways can 3 letters out of 5 distinct letters A [#permalink]
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Expert's post

Hello - Per a PM I received:

for the question :

Ques : In how many ways can 3 letters out of 5 distinct letters A, B, C, D and E be arranged in a straight line so that A and B never come together?

Your Reply : I got 48 as well.

How many distinct letter combos: 5C3 = 10

Each Combo can be arranged in 3! ways 3! = 6

Combos that Include both A & B: 3C1 = 3 In these combos there are not 6 arrangements that work, but only 2 (AxB and BxA) 2*3=6

Combos that don't include both A & B = 7 In these combos every sequence fits criteria: 6*7 = 42

42+6=48

Can you please explain me the part in red

Thanks in Advance!!..

So, working backward from the above, the 42 represents all of the permutations of 3 letters groups from the 5 letters that don't include both A and B (ACD, ADC, BCE, ECB, etc etc).

The only thing left now is to include the 3 letter combos that do include both A and B, but of course don't violate the rule that they cannot be next to each other.

So, per the red, the first thing to do is to see how many combos there are that have both A and B in them...as we already used up 2 of the 3 available slots, we are only looks for 1 more letter. 3C1 = how many ways can we choose 1 letter from the 3 available letters (A & B & ? --> ? = C D or E --> 3 ways or combos).

Next, how many ways can we arrange these groups with both A & B. Per the above red, I now see the "(AxB and BxA)" is a bit confusing (needed a minute to decipher it myself lol). The "x" is just a place holder that can signify C D or E...probably better to use "?" With the original 42 permutations, each combo had 3! arrangements, however with the given constraint that A & B cannot touch, we cannot use 3! to figure out arrangements in the both A & B combos. The only way for A & B not to touch is for them to be on opposite ends. ACB, BCA, ADB, BDA, AEB, BEA = 6. Each combo has 2 arrangements (A?B and B?A, where ? = C D or E)

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