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Theory : Combinations while dealing with similar and dissimilar things

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

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New post 01 Apr 2016, 05:17
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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
3. AD and BC
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

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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New post 15 Jun 2016, 06:26
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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New post 15 Jun 2016, 06:45
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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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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New post 15 Jun 2016, 07:20
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
3. AD and BC
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?

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New post 15 Jun 2016, 07:37
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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
3. AD and BC
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
3) AD..

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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New post 15 Jun 2016, 08:39
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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New post 15 Jun 2016, 09:20
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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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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New post 27 Oct 2016, 00:19
Thank you very much For posting.this helps me a lot. :-D

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

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New post 27 Jan 2017, 07:39
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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New post 28 Aug 2017, 01:30
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] 28 Aug 2017, 01:30
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