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# PERMUTATIONS AND COMBINATIONS-CONFUSION

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Manager
Joined: 17 Jul 2017
Posts: 193

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05 Apr 2019, 00:59
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Hi can anyone help me out?

Say i have to select a group of 2 students from 3 students ,so no of ways of doing it will be:
say A,B,C are students then
AB,BC,CA(since BA same as AB so not considered)
now if if go by combinations
3c1*2c1=6 ways
now here i divide by 2!
but i donot understand why are we diving by 2! as it is combinations only not permumations.please guide
@bunnel sayantanc2k
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Joined: 02 Sep 2009
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05 Apr 2019, 01:15
vanam52923 wrote:
Hi can anyone help me out?

Say i have to select a group of 2 students from 3 students ,so no of ways of doing it will be:
say A,B,C are students then
AB,BC,CA(since BA same as AB so not considered)
now if if go by combinations
3c1*2c1=6 ways
now here i divide by 2!
but i donot understand why are we diving by 2! as it is combinations only not permumations.please guide
@bunnel sayantanc2k

Choosing 2 out of 3, when the order does not matter, would simply be 3C2 = 3.

If you do 3 options for the first slot multiplied by 2 options for the second slot, then the result must be divided by 2!, to get rid of the duplication. Because 3*2 gives AB (selecting A for the first slot and B for the second slot) as well as BA (selecting B for the first slot and A for the second slot).
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Joined: 17 Jul 2017
Posts: 193

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05 Apr 2019, 01:27
Bunuel wrote:
vanam52923 wrote:
Hi can anyone help me out?

Say i have to select a group of 2 students from 3 students ,so no of ways of doing it will be:
say A,B,C are students then
AB,BC,CA(since BA same as AB so not considered)
now if if go by combinations
3c1*2c1=6 ways
now here i divide by 2!
but i donot understand why are we diving by 2! as it is combinations only not permumations.please guide
@bunnel sayantanc2k

Choosing 2 out of 3, when the order does not matter, would simply be 3C2 = 3.

If you do 3 options for the first slot multiplied by 2 options for the second slot, then the result must be divided by 2!, to get rid of the duplication. Because 3*2 gives AB (selecting A for the first slot and B for the second slot) as well as BA (selecting B for the first slot and A for the second slot).

thanks i understood but my qsn is
say i have 8 pairs of siblings and i have to form a group whcich contains 4 kids but no siblings together?
so here i will selct 16c1 * 14c1*12c1*14c1 so this is my all combinations
now i have to divide by 4! too .But why?
its not permuation,so why are we dividing
any how we are jus selecting not arramnging using 16c1 * 14c1*12c1*14c1
Intern
Joined: 05 Apr 2019
Posts: 4

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05 Apr 2019, 05:11
Bunuel THanks a lot, now I understood clearly..
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Joined: 24 Jun 2008
Posts: 1802

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06 Apr 2019, 15:04
vanam52923 wrote:
its not permuation,so why are we dividing

When something is a permutation, you never divide. If you're solving a permutation question, you're putting things in order: "In how many different orders can Amir, Betty and Carlos line up at a ticket window?" is a permutation problem. The answer is 3*2*1 = 6 (we do not divide by anything).

You only divide when the order of things does not matter. In general, if the order of k things doesn't matter, you can count by first pretending order does matter, then dividing by k! at the end. So in your siblings question, the order of the four siblings we choose does not matter -- we aren't picking, say, a President, a Vice-President, a Secretary and a Treasurer. Because the order of our 4 selections does not matter, we can first pretend it does -- we have 16 choices for the first sibling, 14 for the next, 12 for the next and 10 for the final one, so (16)(14)(12)(10) choices in total -- but we then need to divide by 4!, because the order of the four siblings is not supposed to be important.
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Joined: 05 Apr 2019
Posts: 4

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07 Apr 2019, 23:55
IanStewart wrote:
vanam52923 wrote:
its not permuation,so why are we dividing

When something is a permutation, you never divide. If you're solving a permutation question, you're putting things in order: "In how many different orders can Amir, Betty and Carlos line up at a ticket window?" is a permutation problem. The answer is 3*2*1 = 6 (we do not divide by anything).

You only divide when the order of things does not matter. In general, if the order of k things doesn't matter, you can count by first pretending order does matter, then dividing by k! at the end. So in your siblings question, the order of the four siblings we choose does not matter -- we aren't picking, say, a President, a Vice-President, a Secretary and a Treasurer. Because the order of our 4 selections does not matter, we can first pretend it does -- we have 16 choices for the first sibling, 14 for the next, 12 for the next and 10 for the final one, so (16)(14)(12)(10) choices in total -- but we then need to divide by 4!, because the order of the four siblings is not supposed to be important.

IanStewart Great explanation..!
Re: PERMUTATIONS AND COMBINATIONS-CONFUSION   [#permalink] 07 Apr 2019, 23:55
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