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
CAT-holics 
Wonderful post!
Thank You! Asfakul – Sent from my Galaxy Note 2
what if we have to divide 30 guys into 4 teams how to do that
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
Hi. Can you please explain what you mean by the teams being identical and therefore the need to further divide by 2!?
Hi. Can you explain what you mean by the teams are identical and therefore the further need to divide by 2!?
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
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 ?
No, it will be 30C15 (as the teams will now be distinct).
regards
J
Yes…sorry..it should be 30c15…thanks a lot Jš
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?
Because the groups are of the same size and hence indistinguishable.
regards
J
thanks a lot sir š
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
regards
J