How many solutions (a, b, c, d) are there to the equation
a+b+c+d=2007 where a, b, c, d are positive integers?
plz show work. thz.
2007-11-03 11:43:06 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1+1+1+2004
1+1+2 +2003
1+1+3+2002
.......................
1+1+2004+1
So are show 2004 possibilities
This can be repeated 4 times so total is 8016
2007-11-03 11:59:26 · answer #1 · answered by ironduke8159 7 · 0⤊ 1⤋
2006C3 (read 2006 choose 3)
Consider the sequence of 2007 1's:
1,1,1,1,...,1
We choose 3 distinct dividing lines each between some two 1's. Then a will be the sum of all the 1's to the first 1 to the first line, b will be the sum of all the 1's between the 1st line and the 2nd line, and so on. It is evident that a+b+c+d=2007 and number of ways choosing 3 distinct lines is 2006C3
2007-11-03 18:59:51 · answer #2 · answered by moshi747 3 · 0⤊ 1⤋
the highest value that a,b,c or d can be, assuming that they all have to be different, is 2001.
so the calculation is 4! x 2001 = 24 x 2001 = 48024
but if they can be the same values then it would be 4! x 2007 = 48158
2007-11-03 18:52:40 · answer #3 · answered by Anonymous · 0⤊ 1⤋
a = 2007 - b - c- d
b = 2007 - a - c - d
c = 2007 - a - b - d
d = 2007 - a - b - c
2007-11-03 18:49:05 · answer #4 · answered by Fub! 2 · 0⤊ 1⤋