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01 Dec 2003, 12:26
Kudos
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I understand the concept behind Comb. vs. Perm, but sometimes i have problems discerning which one to use in a word problem...for instance:
In how many ways can we choose the 5 out of 35 students that will sit in the first row in MATH250? here you would use comb. becasue order does not matter,
but in this problem, you use perm.:
In how many ways can 5 out of 35 students be seated in the first row in MATH250?
are there trigger words or something that helps you to figure out which to use?
In how many ways can we choose the 5 out of 35 students that will sit in the first row in MATH250? here you would use comb. becasue order does not matter,
but in this problem, you use perm.:
In how many ways can 5 out of 35 students be seated in the first row in MATH250?
are there trigger words or something that helps you to figure out which to use?
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01 Dec 2003, 13:24
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mbartow
I understand the concept behind Comb. vs. Perm, but sometimes i have problems discerning which one to use in a word problem...for instance:
In how many ways can we choose the 5 out of 35 students that will sit in the first row in MATH250? here you would use comb. becasue order does not matter,
but in this problem, you use perm.:
In how many ways can 5 out of 35 students be seated in the first row in MATH250?
are there trigger words or something that helps you to figure out which to use?
In how many ways can we choose the 5 out of 35 students that will sit in the first row in MATH250? here you would use comb. becasue order does not matter,
but in this problem, you use perm.:
In how many ways can 5 out of 35 students be seated in the first row in MATH250?
are there trigger words or something that helps you to figure out which to use?
think if an arrangement is Unique...
is ABCDE different than BCDAE
if you are selecting five from 35...the order in which you select is NOT important... ABCDE is the same as BCDAE..
BUT..
if we are first selecting 5 out of 35 and then arranging them in a row..
we would first use COMBINATIONS to find out in how many ways we can pick 5 out of 35
thats just 35C5 = 35! / 5! * 30! => lets say N ways
now , each of these "N" ways have to arranged in a ROW , which can seat FIVE. lets say ABCDE is one of the selections...
With ARRANGEMENTS, each Arrangement is unique !
so we use permutations , thats 5!
the answer 35! / 5! * 30! * 5! = 35*34*33*32*31 ways
A quicker way :
Any of these 35 students can be seated in the first position in the row
34 students can be seated in the second position
33 students in the third position
32 students in the fourth position
31 students in the fifth position.
The total # of ways = 35 * 34 * 33 * 32 * 31
thanks
praetorian
01 Dec 2003, 14:15
Kudos
Bookmarks
mbartow
I understand the concept behind Comb. vs. Perm, but sometimes i have problems discerning which one to use in a word problem...for instance:
In how many ways can we choose the 5 out of 35 students that will sit in the first row in MATH250? here you would use comb. becasue order does not matter,
but in this problem, you use perm.:
In how many ways can 5 out of 35 students be seated in the first row in MATH250?
are there trigger words or something that helps you to figure out which to use?
In how many ways can we choose the 5 out of 35 students that will sit in the first row in MATH250? here you would use comb. becasue order does not matter,
but in this problem, you use perm.:
In how many ways can 5 out of 35 students be seated in the first row in MATH250?
are there trigger words or something that helps you to figure out which to use?
Think to yourself:
I am picking a "committee" or "group of people" who are free to arrange themselves,
or
am i "lining up" distinguishable objects in a specific order in specific locations.
