find the number of ordered pairs(a,b,c,d) such that abcd=10800?

10800 = $$3^3 * 2^4 * 5^2$$

To find the number of ordered sets, we have to distribute the powers of each of the factors among the four variables.

For the power of 3: a + b + c + d = 3, where a, b, c, d >= 0 => Number of solutions = $$^{3+4-1}C_{4-1} = ^6C_3 = 20$$
For the power of 2: a + b + c + d = 4, where a, b, c, d >= 0 => Number of solutions = $$^{4+4-1}C_{4-1} = ^7C_3 = 35$$
For the power of 5: a + b + c + d = 2, where a, b, c, d >= 0 => Number of solutions = $$^{2+4-1}C_{4-1} = ^5C_3 = 10$$

So, total number of ordered sets = 20 * 35 * 10 = 7000