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PERMUTATIONS AND COMBINATIONSCONFUSION
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05 Apr 2019, 00:59
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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Re: PERMUTATIONS AND COMBINATIONSCONFUSION
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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 sayantanc2kChoosing 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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Re: PERMUTATIONS AND COMBINATIONSCONFUSION
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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 sayantanc2kChoosing 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



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Re: PERMUTATIONS AND COMBINATIONSCONFUSION
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05 Apr 2019, 05:11
Bunuel THanks a lot, now I understood clearly..



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Re: PERMUTATIONS AND COMBINATIONSCONFUSION
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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 VicePresident, 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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Re: PERMUTATIONS AND COMBINATIONSCONFUSION
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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 VicePresident, 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 COMBINATIONSCONFUSION
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