1: A computer locks up if more than 4 applications are running at the same time. If the owner has 20 applications stored in his program file, how many different sets of 4 can he run, so that his computer does not lock up?
2006-08-28 05:39:33 · 8 answers · asked by Matthew B 2 in Science & Mathematics ➔ Mathematics
20 choose 4 (a basic formula for combinations)
The excalmation point means "factorial" i.e. 4! = 4*3*2*1
(20)-C-(4)= (20!) / [(20 - 4)! * (4!)]
=(20!) / [(16)! * (4!)]
=[20*19*18*17*16*15*...*2*1] / [(16*15*...*2*1)*(4*3*2*1)]
=[20*19*18*17] / [4*3*2*1] * [16*15*...*2*1] / [16*15*...*2*1]
=[20*19*18*17] / [4*3*2*1] * [1]
=[5*19*6*17] / [2*1]
=19*17*15
= 4845 sets of 4
2006-08-28 13:58:15 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I will correct WideAwake's answer:
#1 - 20 things taken 4 at a time, order IS important, is 20*19*18*17 or 116,280.
#2 - This assumes that a person is not running the same program again (perhaps with different data). If you can have the same program running 4x, then the combinations are 20*20*20*20 or 160,000.
#3 - If two combinations of the same programs are equivalent, in other words running programs A,B,C,D is the same as running A,D,B,C, then you would divide #1 by 4! or 24, and get 4845.
In math terms, #1 is 20! / 16!, #3 is 20! / (16!)(4!)
2006-08-28 12:54:38 · answer #2 · answered by Anonymous · 0⤊ 0⤋
#1 - 20 things taken 4 at a time, order not important, is 20*19*18*17 or 116,280.
#2 - This assumes that a person is not running the same program again (perhaps with different data). If you can have the same program running 4x, then the combinations are 20*20*20*20 or 160,000.
#3 - If two combinations of the same programs are equivalent, in other words running programs A,B,C,D is the same as running A,D,B,C, then you would divide #1 by 4! or 24, and get 4845.
In math terms, #1 is 20! / 16!, #3 is 20! / (16!)(4!)
2006-08-28 12:47:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(20*19*18*17)/(1*2*3*4 ) = 4845 different sets
2006-08-28 12:46:47 · answer #4 · answered by Dimos F 4 · 0⤊ 0⤋
n=20, k=4. If you are attempting this problem, you know the formula.
2006-08-28 12:45:31 · answer #5 · answered by banjuja58 4 · 0⤊ 0⤋
20 choose 4 = 4 845
(http://www.google.com/search?hl=en&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=20+choose+4&spell=1)
2006-08-28 12:50:57 · answer #6 · answered by mr p 1 · 0⤊ 0⤋
nPr = (n!)/(r!(n - r)!)
20P4 = (20!)/(4!(20 - 4)!)
20P4 = (20!)/(4!(16!))
20P4 = 4845
2006-08-28 20:36:27 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
I presume you meant "different" sets of 4
n! / ((r!*(n-r)!)
2006-08-28 12:47:13 · answer #8 · answered by Gemelli2 5 · 0⤊ 0⤋