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# Theory : Combinations while dealing with similar and dissimilar things

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Math Expert
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Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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01 Apr 2016, 04:17
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Few days back there was a Q on choosing 4 persons in four couple in such a way that NO couple is in the group and then there were Queries WHY there is certain restrictions while finding ways. Also there was a PM on grouping / combinations of dissimilar things in equal groups.

I always ask students interacting with me to know HOW of a thing/process, if TIME permits. Knowing HOW can help you solve otherwise complicated Qs and SAVE time.

Look at the two Q below-

1) what are the ways to distribute 8 people in group of 5 and 3? Ans 8C5 or 8C3..
or say 8 people in groups of 4, 3 and 1? Ans 8C4*4C3*1C1

2) what are the ways to distribute 4 people in group of 2 each?
ans 4C2= 4!/2!2!=6 IS IT SO?
SAY 4 people are ABCD..
groups =>
1. AB and CD
2. AC and BD
WHERE have the REMAINING 3 GONE?
ANS- the other 3 became repetitions of the above three
so ans is 4C2/2!

Similarily if you were dividing 9 in three groups of 3 each, the answer will not be 9C3*6C3 but 9C3*6C3/3!..
since when we made one group of ABC, we simultaneously made DEF and GHI..

Formula-

If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= $$\frac{(xa)!}{x!(a!)^x}$$

lets differentiate between similar and dissimilar things-

No of balls = 4 and number of boxes=3
four cases
1) All balls and All boxes are similar
2) Balls are similar BUT boxes are NOT.
3) Ball are dissimilar and Boxes are similar
4) All balls and All boxes are dissimilar

Lets find the solution for each case-

1) All balls and All boxes are similar

a) All four balls in one box----------------- Ways = 0,0,4 = 1 way
b) 3 in one and 1 in other------------------ ways= 0,1,3 = 1 way
c) 2 in one and 2 in other------------------ ways = 0,2,2 = 1 way
d) 2 in one, 1 in second and 1 in third box-- ways= 1,1,2= 1 way
TOTAL = 4 ways

2) Balls are similar BUT boxes are NOT.
Now all groupings will be similar to above but the boxes are dissimilar, so lets find ways

a) All four balls in one box-- Ways = 0,0,4 = 1 way ; 4 can be any of the box ------------------------------SO TOTAL = 1*3= 3ways
b) 3 in one and 1 in other-- ways= 0,1,3 = 1 way ; the three boxes can be choosen in 3! ways------------SO TOTAL = 1*3!=6 ways
c) 2 in one and 2 in other-- ways = 0,2,2= 1 way ; boxes can be choosen in 3!/2!=3 ways----------------SO TOTAL = 1*3=3 ways
d) 2 in one, 1 in second and 1 in third box-- ways= 1,1,2= 1 way ; boxes can be choosen in 3!/2!=3 ways SO TOTAL = 1*3=3 ways
TOTAL = 3+6+3+3=15 ways

3) Ball are dissimilar and Boxes are similar
Here boxes are the same but the balls are dissimilar

a) All four balls in one box-- Ways = 0,0,4 = 1 way ; since balls are in same box ----------------------------------------------------------TOTAL = 1way
b) 3 in one and 1 in other-- ways= 0,1,3 = 1 way ; balls can be choosen in 4C3 = 4 ways -----------------------------------------------TOTAL = 1*4=4 ways
c) 2 in one and 2 in other-- ways = 0,2,2= 1 way ; balls can be choosen in 4C2/2! = 3 ways (REMEMBER WHAT WE STARTED WITH) TOTAL = 1*3=3 ways
d) 2 in one, 1 in second and 1 in third box-- ways= 1,1,2= 1 way ; balls can be choosen in 4C2 = 4!/2!2!=6 ways-------------------- TOTAL = 1*6= 6 ways
TOTAL = 1+4+3+6 ways = 14 ways

4) All balls and All boxes are dissimilar
MANY Qs belong to this category.
After finding the ways in third, we work on the solution further

a) All four balls in one box= 1way --- box can be any 1 of the three= 1*3 ----------------------------------------------------------------------------------------------- 3 ways
b) 3 in one and 1 in other-- balls can be choosen in 4C3 = 4 ways and Box can be choosen in 3! ways TOTAL = 1*4*6 = ---------------------------------------------24 ways
c) 2 in one and 2 in other-- balls can be choosen in $$\frac{4C2}{2!} = 3$$ ways and Box can be choosen in 3! ways TOTAL = 1*3*6 = ----------18 ways
d) 2 in one, 1 in second and 1 in third box-- ; balls can be choosen in $$4C2 = \frac{4!}{2!2!}=6$$ ways and Box can be choosen in 3! ways TOTAL = 1*6*6 = 36 ways
TOTAL = 3+24+18+36 ways = 81 ways

straight formula here would be 3 can in any of boxes = 3*3*3*3=81

Whenever you look at any Q of distributing items among people or in boxes, the Q will deal with any one of the CASEs mentioned above.
LEARN to differentiate these 4 groups and you may be able to make OTHERWISE a complicated Q LOOK rather easy..

Neither TIME nor SPACE permits further dwelling on the TOPIC, but you can ask any queries if you have
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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15 Jun 2016, 08:20
4
2
nishi999 wrote:
so this EXPLANATION leads us to formula If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= $$\frac{(xa)!}{x!(a!)^n}$$

Hi,

What does the "n" in the formula stand for. Could you illustrate the usage of the formula in an example.

Thanks,[/quote]

Hi,
it is not n but x...

example...
if you are dividing 8 items in 4 groups of 2 items.....
so Normal method = $$\frac{8C2*6C2*4C2}{4!}$$..... divison by 4! is to negate the repetitions..
so $$\frac{8!}{6!2!} * \frac{6!}{4!2!}* \frac{4!}{2!2!} *\frac{1}{4!} = \frac{8!}{2!2!2!2!4!} = \frac{8!}{(2!)^44!}$$......
this is nothing BUT the direct formula we are talking about..
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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15 Jun 2016, 05:26
1
1) what are the ways to distribute 8 people in group of 5 and 3? Ans 8C5 or 8C3..
or say 8 people in groups of 4, 3 and 1? Ans 8C4*4C3*1C1

Hi,

In Number of ways to distribute 8 people in groups of 5 and 3, you have mentioned 8C5 or 8C3, whereas for 8 people in groups of 4, 3 and 1 you mention 8C4*4C3*1C1.
Is this because, after choosing 5 people out of 8, you are left with 3 people and the ways to choose 3 people out of 3 is 1 and hence you only mention 8C5 or 8C3 instead of mentioning 8C5*3C3 or 8C3*5C5.

In regard to 8C4*4C3*1C1, could you confirm the logic applied i.e. After choosing 4 people out of 8, you are left with only 4 more people to choose the next 3 from and hence 4C3. And after that you are left with only 1 person and and hence 1C1. Am i missing something major out here?

Thanks
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Joined: 02 Aug 2009
Posts: 8308
Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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15 Jun 2016, 05:45
2
nishi999 wrote:
1) what are the ways to distribute 8 people in group of 5 and 3? Ans 8C5 or 8C3..
or say 8 people in groups of 4, 3 and 1? Ans 8C4*4C3*1C1

Hi,

In Number of ways to distribute 8 people in groups of 5 and 3, you have mentioned 8C5 or 8C3, whereas for 8 people in groups of 4, 3 and 1 you mention 8C4*4C3*1C1.
Is this because, after choosing 5 people out of 8, you are left with 3 people and the ways to choose 3 people out of 3 is 1 and hence you only mention 8C5 or 8C3 instead of mentioning 8C5*3C3 or 8C3*5C5.

In regard to 8C4*4C3*1C1, could you confirm the logic applied i.e. After choosing 4 people out of 8, you are left with only 4 more people to choose the next 3 from and hence 4C3. And after that you are left with only 1 person and and hence 1C1. Am i missing something major out here?

Thanks

Hi nishi999,
you are absolutely correct in the understanding.....
Also 8C5 = 8C3....

and in 2nd set of groups 4, 3 and 1.... the logic is What you have mentioned, you have 3 to choose from 4, after choosing 4 out of 8..
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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15 Jun 2016, 06:20
2) what are the ways to distribute 4 people in group of 2 each?
ans 4C2= 4!/2!2!=6 IS IT SO?
SAY 4 people are ABCD..
groups =>
1. AB and CD
2. AC and BD
WHERE have the REMAINING 3 GONE?
ANS- the other 3 became repetitions of the above three
so ans is 4C2/2!

Similarily if you were dividing 9 in three groups of 3 each, the answer will not be 9C3 but 9C3/3!..
since when we made one group of ABC, we simultaneously made DEF and GHI..

Hi,

Sorry for bombarding you with the basic level questions, but P & C seems to be killing me. Thanks in advance for bearing with the same.

Concerning your ABCD example, i believe the Number of combinations of 4 diff things taken 2 at a time are - 4C2 = 6. For ABCD they would be AB, AC, AD, BC, BD, CD. In case we had to make groups of 2, then we would divide these 6 by 2, forming 3 groups of 2 each. I am guessing this is what you are trying to state in your example and hence the manner in which you have enunciated it.

What i don't understand is, you state that "Where have the remaining 3 gone" and "Other 3 Become Repetitions of the above". I believe that there is no other 3, as once divided by 2 the answer must be only 3 groups of 2 each i.e. 1. AB & AC 2. AD & BC and 3. BD & CD. Or if divided by 3, the answer would be two groups of 3 groups in them each i.e. 1 group - AB, AC, AD and BC, BD, CD. So where does the question of repetition arise?

Could you enunciate on the same?
Math Expert
Joined: 02 Aug 2009
Posts: 8308
Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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15 Jun 2016, 06:37
1
nishi999 wrote:
2) what are the ways to distribute 4 people in group of 2 each?
ans 4C2= 4!/2!2!=6 IS IT SO?
SAY 4 people are ABCD..
groups =>
1. AB and CD
2. AC and BD
WHERE have the REMAINING 3 GONE?
ANS- the other 3 became repetitions of the above three
so ans is 4C2/2!

Similarily if you were dividing 9 in three groups of 3 each, the answer will not be 9C3 but 9C3/3!..
since when we made one group of ABC, we simultaneously made DEF and GHI..

Hi,

Sorry for bombarding you with the basic level questions, but P & C seems to be killing me. Thanks in advance for bearing with the same.

Concerning your ABCD example, i believe the Number of combinations of 4 diff things taken 2 at a time are - 4C2 = 6. For ABCD they would be AB, AC, AD, BC, BD, CD. In case we had to make groups of 2, then we would divide these 6 by 2, forming 3 groups of 2 each. I am guessing this is what you are trying to state in your example and hence the manner in which you have enunciated it.

What i don't understand is, you state that "Where have the remaining 3 gone" and "Other 3 Become Repetitions of the above". I believe that there is no other 3, as once divided by 2 the answer must be only 3 groups of 2 each i.e. 1. AB & AC 2. AD & BC and 3. BD & CD. Or if divided by 3, the answer would be two groups of 3 groups in them each i.e. 1 group - AB, AC, AD and BC, BD, CD. So where does the question of repetition arise?

Could you enunciate on the same?

Hi,

the point being conveyed is that the moment you form ONE group, the SECOND is automatically made......
so when we are making groups of two in 4 people, normal method is 4C2 and this gives you answer as 6.....
But we get only three groups...
1) AB
2) AC

what happens when we take BC... IT becomes repetition of (3) as BC was automatically formed when AD was made.....
so this EXPLANATION leads us to formula If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= $$\frac{(xa)!}{x!(a!)^n}$$
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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15 Jun 2016, 07:39
so this EXPLANATION leads us to formula If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= $$\frac{(xa)!}{x!(a!)^n}$$[/quote]

Hi,

What does the "n" in the formula stand for. Could you illustrate the usage of the formula in an example.

Thanks,
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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26 Oct 2016, 23:19
Thank you very much For posting.this helps me a lot.
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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27 Jan 2017, 06:39
Anyone to highlight on the differences among the reds?

Balls are similar BUT boxes are NOT
2 in one and 2 in other
Ways = 0,2,2= 1 way; boxes can be chosen in 3!/2!=3 ways, SO, TOTAL = 1*3=------------3 ways

2 in one, 1 in second and 1 in third box
Ways= 1,1,2= 1 way; boxes can be chosen in 3!/2!=3 ways SO TOTAL = 1*3=---------------3 ways

All balls and all boxes are dissimilar
2 in one and 2 in other
Balls can be chosen in 4C2/2!=3 ways and Box can be chosen in 3! ways TOTAL= 1*3*6 =--18 ways

2 in one, 1 in second and 1 in third box
Balls can be chosen in 4C2=6 ways and Box can be chosen in 3! ways TOTAL=1*6*6 =------36 ways
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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28 Aug 2017, 00:30
the fourth case use the calculation that is based on results of the third and the second case. Be alert that groups are different, not same.
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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31 Dec 2018, 10:57
Mahmud6 wrote:
Anyone to highlight on the differences among the reds?

Balls are similar BUT boxes are NOT
2 in one and 2 in other
Ways = 0,2,2= 1 way; boxes can be chosen in 3!/2!=3 ways, SO, TOTAL = 1*3=------------3 ways

2 in one, 1 in second and 1 in third box
Ways= 1,1,2= 1 way; boxes can be chosen in 3!/2!=3 ways SO TOTAL = 1*3=---------------3 ways

All balls and all boxes are dissimilar
2 in one and 2 in other
Balls can be chosen in 4C2/2!=3 ways and Box can be chosen in 3! ways TOTAL= 1*3*6 =--18 ways

2 in one, 1 in second and 1 in third box
Balls can be chosen in 4C2=6 ways and Box can be chosen in 3! ways TOTAL=1*6*6 =------36 ways

Greetings chetan2u

please, can you explain this question?
Although boxes are dissimilar in both cases, we multiplied by 3!/2! in the first, but multiplied by 3! in the second.
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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06 Jul 2019, 20:38
4) All balls and All boxes are dissimilar
MANY Qs belong to this category.
After finding the ways in third, we work on the solution further

d) 2 in one, 1 in second and 1 in third box-- ; balls can be choosen in 4C2=4!2!2!=64C2=4!2!2!=6 ways and Box can be choosen in 3! ways

Question : Why is the selection of balls not considered for the second box? 2C1?

Regards,
Urshilah.
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Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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13 Sep 2019, 03:34
urshi wrote:
4) All balls and All boxes are dissimilar
MANY Qs belong to this category.
After finding the ways in third, we work on the solution further

d) 2 in one, 1 in second and 1 in third box-- ; balls can be choosen in 4C2=4!2!2!=64C2=4!2!2!=6 ways and Box can be choosen in 3! ways

Question : Why is the selection of balls not considered for the second box? 2C1?

Regards,
Urshilah.

Hi urshi

The reason you don't need to select the remaining two balls is because they will be one and one each anyway. You could still "choose" the remaining two by multiplying by 2C1 and 1C1 but to account for the fact they both have the same number of balls, you would need to divide again by 2! which would leave you with the same result as if you had not multiplied by 2C1*1C1 in the first place.

Let me give another example to hopefully help make this a little more clear. Instead of balls and boxes, imagine we have 4 fruits and 3 children:

Donuts: Apple (A), Blueberry (B), Cherry (C), Dates (D)
Children: Micheal (M), Nathan (N), Oliver (O).

Now imagine you have to hand out exactly 2 fruits to one child, and the remaining children get exactly one fruit each. It is up to you how to distribute (this is the same as 4(d)).

I like to visualize this problem in two phases:

(1) Irrespective of who gets what, I need to decide which fruits will be handed out in a group of 2 and which will be handed out individually. When you think about it, all i need to do is select which two to give together, because whatever is left over will automatically be given individually. I do not need to worry at this stage about the ORDER they will be given out because I am only worrying about the UNIQUE COMBINATIONS at this stage. The order will be determined by the next step. For example {AB, C, D} is identical to {AB, D, C}. This is why I do not need to do 2C1 (or if I do multiply by 2C1 I need to divide by 2! to acknowledge there are 2 identical groups).

(2) Now once I have decided the fruit groupings I need to decide in which order the children will line up to collect their fruits. There are 3! ways this could happen.

To make this point even clearer, imagine now that I had 5 fruits and 4 children, and I wanted to distribute 2 fruits to one child and the remaining 3 to one child each {2,1,1,1}.

Now I could go along with what you suggested and calculate the combinations by 5C2*3C1*2C1*1C1 however once again there are 3 identical groups of one fruit each so I would need to divide by 3! .
The other way to think about this as mentioned above is that you only really need to choose which 2 go together (5C2) because the rest will automatically be 1 each. (Remember at step 1 we are only calculating COMBINATIONS. We will account for the fact that each of these are different fruits in step (2)).

Hope this helps
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Re: Theory : Combinations while dealing with similar and dissimilar things  [#permalink]

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23 Feb 2020, 21:32
Hi,

straight formula here would be 3 can in any of boxes = 3*3*3*3=81

Could you kindly explain how you arrived at this when the question initially given was 4 balls and 3 boxes? Also when is it applicable to use the formula xa!/x!(a!)^x

Re: Theory : Combinations while dealing with similar and dissimilar things   [#permalink] 23 Feb 2020, 21:32
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