In how many ways can 3 letters out of five distinct 5 distinct letters

Here's the solution.

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!

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

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.

Total = 60-18=42.

Its 42

5P3-3!*3=42

Hi,

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.

I also got 42.

But...That's a poorly worded question. Nothing worded like this will ever show up on a real GMAT or any of the books published in the US.

This is my take on this:

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.

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

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)

42 + 6 = 48

Hope this helps.

"arrange 3 letters out of 5 such that A and B are there and together in 3*2!*2 = 12 ways


Hi Karishma ,

Will you please explain how you got 3 * 2! *2 ?

Thanks in advance

You have already selected A and B. Now select one other letter out of the remaining three in 3C1 = 3 ways.
So you have A, B and one more letter. A and B should be together forming a group. So you need to arrange AB and the third letter. This can be done in 2! ways.
But Ab can also be arranged in 2 ways: AB and BA.

Hence, total arrangements = 3 * 2! * 2

Hello Karishma ,

I have seen answer to this problem somewhere as (5C3-3C1) * 3!

could you please help explain 3C1 part?

Thanks in advance.

VeritasPrepKarishma : I understand the logic of 5*4*3 with the total permutation = 60 - 12 = 48. However, if i were to go to the logic directly, there are 5 no's. I was placing the numbers directly, So 5(all 5 characters) *3 (3 being without b) *3 (the remaining characters) = 45.

Am i missing something?

When you say the first position can be taken by any 5 characters, the first position can be taken by A/B/C/D/E. Say the first position is taken by E

E ____ ____

Now you are ignoring B in the second position. That is not correct. You can have E B ___ provided you don't have A in the last position. So something like E B C is possible though you are not counting it.

Remember that whenever you have "A and B should not be together", putting them together and subtracting out from the total will be far easier to do.

5C3 is the number of ways in which you select any 3 letters out of the 5. This will include cases in which both A and B are selected along with 1 other letter.
3C1 is the number of ways in which you can select A, B and one other letter from the remaining 3 letters.
3! is the number of ways in which you arrange these 3 letters.

So you are selecting 3 letters out of 5 such that A and B are not selected together. The problem here is that you missed cases 6 cases: ACB, ADB, AEB, BCA, BDA, BEA - i.e. all cases in which A and B are selected but not placed together. These are acceptable.

(5C3-3C1) * 3! gives you 42 only. The answer is 48.

VeritasPrepKarishma : Yeah i got the mistake. But i'm looking at alternative methods to reach the answer.

Any other effective method? Kindly do share.