Question 837677: A family has 12 members. In how many ways can six family members be seated in a row so that their ages increase from left to right?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe the answer will be 12C6 which is equal to 12! / (6! * 6!).
this is the combination formula where order doesn't matter.
why did i choose this instead of the permutation formula where order does matter?
my reasoning is as follows:
you want a group of 6 people who are unique so that you can then arrange them in order of increasing age.
a permutation would give you the same people in different orders.
this, i believe, is not what you want.
a small example should explain the logic.
suppose you have 4 family members and want to know how many ways 3 family members can be seated in a row so that their ages increase from left to right.
well, once you pick the 3 people there is only one way in which they can be seated with their ages in increasing order.
so you need to pick the individuals that will all be unique in that group first (this is the combination formula of 4C3 where order doesn't matter), and then you want to arrange them in order of increasing age.
let's look at the numbers using 3 out of 4 family members and you should be able to see why the combination formula was used here, and not the permutation formula.
4C3 = 4! / (3! * 1!) which is equal to 4 possible arrangements where order doesn't matter. the combination formula gives you groups where the same 4 members are only in one group and not in any other group.
let's number the 4 people by age with number 1 being the youngest and number 4 being the oldest. the actual age doesn't really matter since number 2 is older than number 1 and number 3 is older than number 2 and number 4 is older than number 3.
so we figure out how many unique groups we can get where order doesn't matter.
those groups are:
1,2,3
1,2,4
1,3,4
2,3,4
there are no other possible combinations where order doesn't matter because these are the only possible groups of 3 where all members or the same group will not be round in any other group of 3.
now that we have these groups, there is only one way they can be arranged by increasing age where the youngest is first and the oldest is last.
that's the order shown.
now let's look at what we would have gotten if we had used permutation instead.
with permutations, order does matter, so the same persons can be formed in several groups where order does matter.
the permutation formula, in this case, would be 4! / 1!.
the result of that is 4*3*2*1 = 24 possible groups.
those groups would be:
1,2,3 1,3,2 2,1,3 2,3,1 3,1,2 3,2,1
1,2,4 1,4,2 2,1,4 2,4,1 4,1,2 4,2,1
1,3,4 1,4,3 3,1,4 3,4,1 4,1,3 4,3,1
2,3,4 2,4,3 3,2,4 3,4,2 4,2,3 4,3,2
you can see that there are multiple groups where the same 3 people are in each group. 6 groups can be formed from the people numbered 1,2,3. 6 groups can be formed from the people numbered 1,2,4. 6 with 1,3,4. 6 with 2,3,4.
in the first group, 1,2,3 is the only order that matters.
in the second group, 1,2,4 is the only order that matters.
in the third group, 1,3,4 is the only order that matters.
in the fourth group, 2,3,4 is the only order that matters.
all those other possible arrangements that you got by using the permutation formula are superfluous in this case.
this is why i believe the combination formula is the one that you want.
applying that combination formula to your problem, you get:
12C6 is equal to 12! / (6! * 6!) which is equal to 924.
that's your answer.
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