A. The number of solutions to the equation x1 + x2 + x3 + x4 + x5 + x6 = 29, where xi is a nonnegative integer, is: a) 22C6, b) 29C5, c) 23C5, and d) 6C122C5 - 1 - 6C114C5.
B.
a) Since xi>1, we can subtract 2 from each xi, which gives us the equation x1' + x2' + x3' + x4' + x5' + x6' = 17, where xi' = xi - 2. Then, we can use stars and bars to get the solution as (17 + 6 - 1)C(6 - 1) = 22C5.
b) Since x2>1, x5>5, and x6>0, we can subtract 2 from x2, subtract 6 from x5, and subtract 1 from x6, which gives us the equation x1 + x2' + x3 + x4 + x5' + x6' = 20, where xi' = xi - 2 for i = 2, 5, and xi' = xi - 1 for i = 6. Then, we can use stars and bars to get the solution as (20 + 6 - 1)C(6 - 1) = 29C5.
c) Since x3≥5, we can subtract 5 from x3, which gives us the equation x1 + x2 + x3' + x4 + x5 + x6 = 24, where x3' = x3 - 5. Then, we can use stars and bars to get the solution as (24 + 6 - 1)C(6 - 1) = 23C5.
d) To find the number of solutions where xi<8 for all i and xi>8 for at least one i, we can subtract 1 from x5, which gives us the equation x1 + x2 + x3 + x4 + x5' + x6 = 21, where x5' = x5 - 1. Then, we can use stars and bars to get the total number of solutions where xi<8 for all i as (21 + 6 - 1)C(6 - 1) = 26C5. To find the number of solutions where xi<8 for all i and xi>8 for at least one i, we can subtract 9 from xi for the i that is greater than 8 and subtract 1 from x5, which gives us the equation x1 + x2 + x3 + x4 + x5' + x6' = 12, where x5' = x5 - 1 and x_i' = x_i - 9 for i>8. Then, we can use stars and bars to get the total number of solutions where xi<8 for all i as (12 + 6 - 1)C(6 - 1) = 16C5. Therefore, the number of solutions where xi<8 for all i except xi>8 for at least one i is 26C5 - 16C5 = 6C122C5 - 1 - 6C114C5.
Note: In all cases, we use the stars and bars technique to count the number of nonnegative integer solutions to an equation of the form x1 + x2 + ... + xn = k. The answer