Question

The total number of ways in which a beggar can be given at least one rupee from four 25-paisa coins, three 50-paisa coins and 2 one-rupee coins, is

• A. 54
• B. 53
• C. 51
• D. none of these

Answer 1

Answer: (A)

54

Let's denote 25 paise with x,50 paise with y and 1 rupee with z.
Hence we have $(x^0+x^1+x^2+x^3+x^4)(y^0+y^1+y^2+y^3)(z^0+z^1+z^2)$
hence the total no. of ways in which a beggar can be paid =$5\times 4\times 3=60$
But for the beggar getting at least 1 rupee we have to subtract the cases when he gets 25 paise or 50 paise or 75 paise.
Hence no. of ways of getting 25,50 and 75 paise is :
25 or
25 25 or
50 or
25 50 or
50 25 or
25 25 25
So we have to remove these $6$ cases.
Therefore, the rest no. of ways $=60-6=54$
Hence, option 'A' is correct.