Number of solutions of an inequality

You can visualize this problem as choosing from 6 numbers and five equalities. For example you can choose any four numbers from 0 to 5 and two equal tos. Hence answer is 11 C 6

Please reformulate the problem to make it clearer. We're no looking for pairs, but for ordered six-tuples $(a_1, a_2,a_3,a_4,a_5,a_6)$

The minimum value of the sequence is 0 and the maximum value can be 5. So we can view the sequence as a path from (1,0) to (6,5) Now any path from (1,0) to (6,5) will represent a valid sequence, as (1, a1) (2, a2) (3, a3)(4,a4)(5,a5)(6,a6) thus our sequence (a1,a2, a3, a4, a5, a6) can be represented by a path from (1, 0) to (6,5) consisting of only "Horizontal(represented by H)" or "Upward(represented by U)" move. i.e. any permutation of 6 H's and 5 U's will represent a valid path from (1,0) to (6,5) and thus can be our solution. So Ans = No. of permutation of (HHHHHHUUUUU) = 11! / (6! * 5!) = 462 Let us see a example for clarification. say, (a1, a2, a3, a4, a5, a6) = (1, 2, 2, 3, 4, 4) then We can View it as a path from (1, 0) to (6, 4) as (1, 1) (2, 2) (3, 2) (4, 3) (5, 4) (6, 4).