A). xi>1fori=1,2,3,4,5,6?
(2+x1)+(2+x2)+(2+x3)+(2+x4)+(2+x5)+(2+x6)=29 x,𝑖 ≥ 0 for i=1, 2, 3, 4, 5,6
x1+x2+x3+x4+x5+x6=17
C(n+k-1, k-1) or C(k+n-1, n)
C(22,5)= 26334
B).x1≥1,x2≥2,x3≥3,x4≥4,x5>5,and x6≥6?
(1+x1)+(2+x2)+(3+x3)+(4+x4)+(5+x5)+(6+x6)=29 x,𝑖 ≥ 0 for i=1, 2, 3, 4, 5,6
x1+x2+x3+x4+x5+x6=8
C(n+k-1, k-1) or C(k+n-1, n)
C(13,5)=1287
C).x1≤5?
First we count the number of solutions with x1 ≥ 6 (note: because we are assuming x1 to be an integer, x1 ≥ 6 is an appropriate form of the logical negation to x1 ≤ 5.).
This is the same as the number of non-negative solutions to y1 + x2 + x3 + x4 + x5 + x6 = 23, that is, C(28, 5).
Then we subtract this from the number of solutions to the original equation. The answer is C(34, 5) − C(28, 5).=179976
D).x1<8 and x2>8
) We now must combine the tricks from the previous items: By letting y2 = x2 − 9,
we see that we want to count the number of solutions to x1 + y2 + x3 + x4 + x5 + x6 = 20
=C(25, 5)
for which x1 < 8. Now we count the complimentary set, so solutions to x1 + y2 + x3 + x4 + x5 + x6 = 20 for which x1 ≥ 8.
Letting y1 = x1 − 8, we see that this is the same as the number of solutions to y1 + y2 + x3 + x4 + x5 + x6 = 12,
which is C(17, 5).
Now we have to not forget that we just counted the compliment of the set we actually care about.
The answer is C(25, 5) − C(17, 5).=46962
