# of ways

[mehrasa]

1) we have 5 people and want to form a 3 member groups.. in how many ways we can form a group of 3 in a way that Ann is in all groups...

To me, the answer is 12... we have 3 places ---, for the first place we put Ann, second place, 4 people out of 5 (except for Ann) and third place 3 people can be seated...also, the order is not important.. then: 4*3*1= 12

2) we have 5 members of a council, G,A,M,R,T.. in how many ways committee can be seated in a way that R and T sit always next to each other?

for solving this problem, we need to consider R and T as one member so we have {G,A,M, {R-T}}
then the total ways is 4! and the # of orders, R and t can position is 2!--> the answer is 4!*2!(this is the part i can not understand, why we multiply this two while we know that when we have indistinguishable objects (R and T) we need to say 5!(total)/ 2!

could you please clarify me.. thnx

[pemdas]

when you arrange by 5! you violate "sit together" condition, this is when R-T are clued and seat together 4!

when factoring 4! is multiplied by 2! it means that 4! ways of clued arrangement R-T is possible in 2! ways, that is in reversed orders T-R too

[mehrasa]

HEy Pemdas

thnx for ur clarification...good point.. what do u think abt Q1, do u think I am on the right track?

[pemdas]

with 3 people, it's easier to know that we select them in 1 way (order doesn't matter)
with 4 people, we form three-person groups in C(4,3)=4 ways, that is one additional person forms three new groups with three other people (out of four) and three old people have their one group (where Ann was sitting). The chance that Ann will sit in new groups of four people is 2 more additional ways (in total 3 out of 4)
with 5 people there are 3 more additional new ways on top of 3 existing ways
>>> in total 6 ways

@mehrasa, you say the order is not important but then you permute Ann, next person (4 out of 5), etc.
would you apply step by step logic above?

3 people ==> three-person groups >>> 1 way
4 people ==> three-person groups >>> 1 existing way + 3 new ways
...

1) we have 5 people and want to form a 3 member groups.. in how many ways we can form a group of 3 in a way that Ann is in all groups...

To me, the answer is 12... we have 3 places ---, for the first place we put Ann, second place, 4 people out of 5 (except for Ann) and third place 3 people can be seated...also, the order is not important.. then: 4*3*1= 12

[mehrasa]

with the logic u mentioned
5 people ==> 1 existing way + 6 new ways = 7
do u mean the answer is 7?

another Q, are u sure that when I use --- this is method for permutation? I am really confused now

[pemdas]

plz forget arrangements out of three, four etc. these were examples to show way

see drawing below (if all else fails, explain by drawing on the paper and using examples)

with the logic u mentioned
5 people ==> 1 existing way + 6 new ways = 7
do u mean the answer is 7?

another Q, are u sure that when I use --- this is method for permutation? I am really confused now