Sherlock Homes and Dr. Watson have to travel from Rajeev Gandhi chowk to Indira
Gandhi International Airport via the metro. They have enough coins of 1, 5, 10 and 25
paise. Sherlock Homes agree to pay for Dr. Watson only if he tells all the possible
combinations of coins that can be used to pay for the ticket.
How many combinations are possible if the fare is 50 paise ?	Answers.
A) 52	B) 49	C) 45	D) 44	E) 42
Let the number of 1, 5, 10 and 50 paise coins required are x, y, z and w respectively.   So when paid total amount is: x + 5y + 10z + 25w, which is equal to the fare paid.   So, x + 5y + 10z + 25w = 50   The number of non negative integral solution of this equation will be equal to the   number of ways the fare can be paid.   x + 5y + 10z + 25w = 50   In this equation w can be either 0, 1 or 2   Case 1: When w is 2   No other coins are required to pay   So there is only one way to pay: (0, 0, 0, 2)   Total no of ways : 1   Case 2: When w is 1   x + 5y + 10z = 25   In this case z can have only three values possible: 0, 1 or 2   Case I: When z is 0   x + 5y = 25   possible values are: (25, 0, 0, 1), (20, 1, 0, 1), (15, 2, 0, 1), (10, 3, 0, 1),   (5, 4, 0, 1), (0, 5, 0, 1)   Case II: When z is 1   x + 5y = 15   possible values are: (15, 0, 1, 1), (10, 1, 1, 1), (5, 2, 1, 1), (0, 3, 1, 1)   Case III: When z is 2   x   + 5= 
<gadget id="calculator">  + 5</gadget>
<output>5</output>
5   possible values are: (5, 0, 2, 1), (0, 1, 2, 1)   Total no of ways : 12   Case 3: When w is 0   x + 5y + 10z= 50   In this case z can have only three values possible: 0, 1, 2, 3, 4 or 5   Case I: When z is 0   x + 5y = 50   possible values are: (50, 0, 0, 0), (45, 1, 0, 0), (40, 2, 0, 0), (35, 3, 0, 0),   (30, 4, 0, 0), (25, 5, 0, 0), (20, 6, 0, 0), (15, 7, 0, 0), (10, 8, 0, 0), (5, 9, 0, 0),   (0, 10, 0, 0)   Case II: When z is 1   x + 5y = 40   possible values are: (40, 0, 1, 0), (35, 1, 1, 0), (30, 2, 1, 0), (25, 3, 1, 0),   (20, 4, 1, 0), (15, 5, 1, 0), (10, 6, 1, 0), (5, 7, 1, 0), (0, 8, 1, 0)   Case III: When z is 2   x + 5y = 30   possible values are: (30, 0, 2, 0), (25, 1, 2, 0), (20, 2, 2, 0), (15, 3, 2, 0),   (10, 4, 2, 0), (5, 5, 2, 0), (0, 6, 2, 0)   Case IV: When z is 3   x + 5y = 20   possible values are: (20, 0, 3, 0), (15, 1, 3, 0), (10, 2, 3, 0), (5, 3, 3, 0), (0,   4, 3, 0)   Case V: When z is 4   x + 5y = 10   possible values are: (10, 0, 4, 0), (5, 1, 4, 0), (0, 2, 4, 0)   Case VI: When z is 5   1 2 3 4 5 6 7 8 9   x + 5y = 0   possible values are: (0, 0, 5, 0)   Total no of ways : 36   So total no of ways = 1 + 12 + 36= 
<gadget id="calculator">1 + 12 + 36</gadget>
<output>49</output>
49   ANSWER:49
<result>B</result>
Sherlock Homes and Dr. Watson have to travel from Rajeev Gandhi chowk to Indira
Gandhi International Airport via the metro. They have enough coins of 1, 5, 10 and 25
paise. Sherlock Homes agree to pay for Dr. Watson only if he tells all the possible
combinations of coins that can be used to pay for the ticket.
How many combinations are possible if the fare is 50 paise ?