Let N be the number of ways of distributing 52 identical balls into 4 distinguishable boxes such that no box is empty and the difference between the number of balls in any two of the boxes is not $a$ multiple of 6. If N = 100a + b, where a, $b$ are positive integers less than 100, find $a$ + b.

Step 1: Let ith box has 6i + i balls where i, i in w and i \le 5. Also, all i’s are distinct.

Step 2:  6(1 + 2 + 3 + 4) + (1 + 2 + 3 + 4) = 52 and 1 + 2 + 3 + 4 in {6, 7, 8, 9, 10, 11, 12, 13, 14}

Step 3: Hence, only one possibility is there

Step 4: 1 + 2 + 3 + 4 = 7 …(i)

Step 5: and 1 + 2 + 3 + 4 = 10 …(ii)

Step 6: Equation (ii) has only three solution sets, which are (5, 4, 1, 0), (5, 3, 2, 0) and (4, 3, 2, 1).

Step 7: Equation (i) has total 10C3 solutions and 4C1  9C2 solutions when any i is zero.

Step 8:  

Step 9: So, N  3( 10C3 ·(4!)) – 2 4C1 · 9C2 (3!)  6912

Step 10: => $a$ = 69 and $b$ = 12.