Think of two people. You could arrange them in 2! = 2 ways normally right? But if they are identical, then you could arrange them in only 1 way. Similarly 4 people could be arranged in 4! ways but if two of them are identical then 4!/2! ways. So identical means we lose a certain, let’s call it a degree of freedom….hence we end up having to divide by 2! here. Do the above case of 4 people in two teams manually and convince yourself, then try to make 6 people into two groups of 3 (by manual listing) and see for yourself how many ways are possible
Sir in the last tennis question just a clarification!
the division of 2! ,is it because of selecting C first and B then (previous case being B first and C second) or the groups as a whole is not distinguishable because of 2 members?
Hi J, For identical groups, like the 1st problem, how do we list the possibilities fast? Is there any specific methodology you follow? This is because it gets cumbersome in the exam if the possibilities are more than 10.
Thanks.
Always do it in a structured manner. So for example as you see I started at the extreme case of 700, then went to cases with 6, then 5 and so on. It reduces the chance of error when doing fast. Having said which, the exam is unlikely to give you an enumeration-based question like this with dozens of possibilities; at max 12 or 15 is realistically likely (historically, even less). Anything larger should probably be left anyway š
Wonderful post!
Thank You! Asfakul – Sent from my Galaxy Note 2
By Asfakul Islam on May 23, 2013 at 8:21 AM
what if we have to divide 30 guys into 4 teams how to do that
By ankit on October 29, 2013 at 4:57 PM
Depends on the size of the teams….teams of different sizes will be treated as distinct by default, same size, it needs to be specified….
regards
J
By catcracker on October 29, 2013 at 5:40 PM
Hi. Can you please explain what you mean by the teams being identical and therefore the need to further divide by 2!?
By kimtre21 on July 29, 2015 at 11:31 AM
Hi. Can you explain what you mean by the teams are identical and therefore the further need to divide by 2!?
By kimtre21 on July 29, 2015 at 11:32 AM
Think of two people. You could arrange them in 2! = 2 ways normally right? But if they are identical, then you could arrange them in only 1 way. Similarly 4 people could be arranged in 4! ways but if two of them are identical then 4!/2! ways. So identical means we lose a certain, let’s call it a degree of freedom….hence we end up having to divide by 2! here. Do the above case of 4 people in two teams manually and convince yourself, then try to make 6 people into two groups of 3 (by manual listing) and see for yourself how many ways are possible
regards
J
By catcracker on July 29, 2015 at 4:12 PM
Hi J,
Very useful concepts. One doubt,
What if 2teams of 15 each, but they have different jerseys say blue and green. Will it be 15c2 ?
By veer on June 20, 2016 at 11:13 AM
No, it will be 30C15 (as the teams will now be distinct).
regards
J
By catcracker on June 20, 2016 at 1:56 PM
Yes…sorry..it should be 30c15…thanks a lot Jš
By veer on June 20, 2016 at 4:12 PM
Sir in the last tennis question just a clarification!
the division of 2! ,is it because of selecting C first and B then (previous case being B first and C second) or the groups as a whole is not distinguishable because of 2 members?
By hari on September 15, 2016 at 11:16 PM
Because the groups are of the same size and hence indistinguishable.
regards
J
By catcracker on September 16, 2016 at 11:31 AM
thanks a lot sir š
By hari on September 18, 2016 at 6:39 PM
Hi J, For identical groups, like the 1st problem, how do we list the possibilities fast? Is there any specific methodology you follow? This is because it gets cumbersome in the exam if the possibilities are more than 10.
Thanks.
By beezgeek on October 29, 2016 at 8:15 PM
Always do it in a structured manner. So for example as you see I started at the extreme case of 700, then went to cases with 6, then 5 and so on. It reduces the chance of error when doing fast. Having said which, the exam is unlikely to give you an enumeration-based question like this with dozens of possibilities; at max 12 or 15 is realistically likely (historically, even less). Anything larger should probably be left anyway š
regards
J
By catcracker on October 30, 2016 at 11:43 AM