Question

1. How many 6 digit strings have a sum of 35? in a digit string the first digit can be zero

2. In how many ways can one arrange the set {A,B,C,D,E} if

E can not be on either end of the string

A must be in an even position

B must be in an odd position (Solve by direct method and P.I.E.)

Answer 1

2)Case 1: If "E" is at 2nd position, for A, we have only one choice, i.e. 4th position, for "B", we have 3 choices, 1st, 3rd or 5th position. Rest of 2, C and D can be arranged in 2! Ways. So no of ways = 1•1•3•2 = 6

CaseII: If "E" is at 3rd position, for A, we have only two choices, i.e. 2nd and 4th position, for "B", we have 32 choices, 1st and 5th position. Rest of 2, C and D can be arranged in 2! Ways. So no of ways = 1•2•2•2 = 8

Case III: If "E" is at 4th position, for A, we have only one choice, i.e. 2nd position, for "B", we have 3 choices, 1st, 3rd or 5th position. Rest of 2, C and D can be arranged in 2! Ways. So no of ways = 1•1•3•2 = 6

Total no of ways = 6+8+6 = 20