# 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)

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.

${2}^{7} = 128$

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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

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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 $n = \text{population", k="picks}$

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

${C}_{3 , 1} {C}_{4 , 1} = 3 \times 4 = 12$

$4 + 6 + 4 + 3 + 12 = 29$
$128 - 1 - 29 = 98$