combination
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Help with a combination problem please. I narrowed it down to C or E but I had to move on. Also i don't know how to get C by using an actual method, i sorta had to use 3 people as an example and counted. If anyone can go about this methodologically it'd be a great help thanks!
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so i'll assume that you got 120 by multiplying 5 x 4 x 3 x 2 x 1. this would be the correct answer if the 5 people were seated in a row, so that every possible ordering would be different from every other possible ordering.
however, the people are seated around a circular table, so that certain orderings can be rotated to produce other, ostensibly "different" orderings.
for instance, let's arbitrarily call one chair the "head" of the table. if the ordering, starting at the head of the table and going, say, clockwise, is ABCDE, that's indistinguishable from BCDEA (or CDEAB, or DEABC, or EABCD).
so, there are going to be fewer than 120 orderings that are actually distinct.
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2 ways of solving the problem:
(1) divide by a redundancy factor:
realize that, because of the aforementioned rotation, every arrangement will be repeated 5 times if you count 5x4x3x2x1 = 120 possibilities.
because every distinct arrangement is actually counted five times, divide by five to find the number of truly distinct arrangements: 120 / 5 = 24.
(2) fix one chair:
you can ensure that each ordering is genuinely different by fixing the seating place of one of the five people. (this way, you know that your solutions can't be rotated into any of the other possible solutions, because the "fixed" person would wind up in the wrong place.)
so, let's say "A" must be listed first.
in this case, then, you're only choosing the seats for the remaining 4 people (B, C, D, E). this can be done freely, so that the number of arrangements is 4 x 3 x 2 x 1 = 24.
however, the people are seated around a circular table, so that certain orderings can be rotated to produce other, ostensibly "different" orderings.
for instance, let's arbitrarily call one chair the "head" of the table. if the ordering, starting at the head of the table and going, say, clockwise, is ABCDE, that's indistinguishable from BCDEA (or CDEAB, or DEABC, or EABCD).
so, there are going to be fewer than 120 orderings that are actually distinct.
--
2 ways of solving the problem:
(1) divide by a redundancy factor:
realize that, because of the aforementioned rotation, every arrangement will be repeated 5 times if you count 5x4x3x2x1 = 120 possibilities.
because every distinct arrangement is actually counted five times, divide by five to find the number of truly distinct arrangements: 120 / 5 = 24.
(2) fix one chair:
you can ensure that each ordering is genuinely different by fixing the seating place of one of the five people. (this way, you know that your solutions can't be rotated into any of the other possible solutions, because the "fixed" person would wind up in the wrong place.)
so, let's say "A" must be listed first.
in this case, then, you're only choosing the seats for the remaining 4 people (B, C, D, E). this can be done freely, so that the number of arrangements is 4 x 3 x 2 x 1 = 24.
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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IMO 24.
I wud have used the formula for circular arangement = (n-1)!
there fore (5-1)!= 24.
let me know the OA.
I wud have used the formula for circular arangement = (n-1)!
there fore (5-1)!= 24.
let me know the OA.
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Even I got 4! or 24 as the answer, using the combination for a circle theory. In a circle one person or seat has to be fixed for assigning seats to remaining 4 people.
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the OA is 24, I didn't know there was a circular formula for it. lol
This'll come in handy for my actual test in just 2 hours. Thanks for the help everyone.
This'll come in handy for my actual test in just 2 hours. Thanks for the help everyone.