Find the total number of ways in which a beggar can be given at least one rupee from four 25paise coins,three 50paise coins & 2 one rupee coins ? (1rupee=100paise)

1 Answer

Answer:

98

Explanation:

There are nine coins in total. I can choose to give one, some, or all of the coins to the beggar. Since each choice for each coin is a yes/no, I can use #2^x# to figure out the total number of combinations.

#2^7=128#

~~~~~

To check this, let's say we have 3 coins called A, B, C. We should have #2^3=8# ways to give the coins:

A
B
C
AB
BC
AC
ABC
none

~~~~~

How many ways out of the 128 makes a rupee?

One way to do this is to count the number directly and the other is to figure out how many ways we don't make a rupee and subtract that from 128. I think I'll approach this by not going over 1 rupee.

What are the different coin combinations that will keep us under? Each of these, being a combination, will fall under this general equation:

#C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"#

1 X 25 paise
2 X 25 paise
3 X 25 paise
1 X 50 paise
1 X 50 paise + 1 X 25 paise

We can evaluate each of these:

1 X 25 paise

#C_(4,1)=(4!)/((1!)(4-1)!)=(4!)/(3!)=24/6=4#

2 X 25 paise

#C_(4,2)=(4!)/((2!)(4-2)!)=(4!)/((2!)(2!))=24/4=6#

3 X 25 paise

#C_(4,3)=(4!)/((3!)(4-3)!)=(4!)/((3!)(1!))=24/6=4#

1 X 50 paise

#C_(3,1)=(3!)/((1!)(3-1)!)=(3!)/((1!)(2!))=6/2=3#

1 X 50 paise + 1 X 25 paise

#C_(3,1)C_(4,1)=3xx4=12#

Let's add them up:

#4+6+4+3+12=29#

Finalizing up

We start with 128 ways to hand out coins. We need to subtract 1 to account for not having any coin, then another 29 for handing out coins that are insufficient to reach 1 rupee. This leaves us with:

#128-1-29=98#