Question on Probability of Coins of different Metals

Topic : Probability
Question : I have 20 coins . I have 1.00 I do not have 20 nickels. what do i have?

Answer :
By trial and Error Method:
Given that not all are nickels, so some can be.
cent nickel dimes quarter
1 5 10 25
number 10 5 4 1 20
amount 10 25 40 25 100

So, it would be 1 quarter, 4 dimes, 5 Nickels and 10 Cents/Pennies, all put together makes 1 dollar

Calculation :
Consider the C as number of cents, D for number of dimes, N for nickels and Q for quarters. Then we have the following equations from the given. (Also to note N not equal to 20)
C + D + N + Q = 20…………………………Eqn1
1C+10D+5N+25Q = 100…………………Eqn2

Eqn2 – Eqn1 gives
9D + 4N +24Q = 80 (sending 9D to the right)
4N+24Q=80-9D (dividing throughout by 4)
N+6Q=20-9/4*D
Since all the unknowns are whole numbers and cannot be negative, we get the following cases:
D can be 0 or 4 or 8
Let us consider each case separately:---
Main Case 1 : D = 0
Then N + 6Q = 20 -0 = 20
N = 20 – 6 Q
This gives two further cases:: Q can be 1 , 2 or 3

Sub case 1 :
Q = 1
Then N = 20-6 =14
Q = 1 , N = 14, D = 0 and c = 5
( checking : 1 (5) + 10(0) + 5( 14) + 25(1) = 5+0+70+25 =100 )
Solution :::: c = 5 , D = 0, N = 14 and Q = 1

Sub case 2 :
Q = 2
Then N = 20-12 =8
Q = 2 , N = 8, D = 0 and c = 10
( checking : 1 (10) + 10(0) + 5( 8) + 25(2) = 10+0+40+50 =100 )
Solution :::: c = 10 , D = 0, N = 8 and Q = 2

Sub case 3 :
Q = 3
Then N = 20-18 =2
Q = 3 , N = 2, D = 0 and c = 15
( checking : 1 (15) + 10(0) + 5(2) + 25(3) = 15+0+10+75 =100 )
Solution :::: c = 15 , D = 0, N = 2 and Q = 3

Main Case 2 : D = 4
Then N + 6Q = 20-9 = 11
N = 11- 6Q
This gives two further cases :: Q can be 0 or 1

Sub case 1 : Q = 0
Then N = 11
So D = 4 and C will be 5
( checking : 1 (5) + 10(4) + 5(11) + 25(0) = 5+40+55+0 =100 )
Solution :::: c = 5 , D = 4, N = 11 and Q = 0

Sub case 2
Q = 1
Then N = 5
So D = 4 and C will be 10
( checking : 1 (10) + 10(4) + 5( 5) + 25(1) = 10+40+25+25 =100 )
Solution :::: c = 10 , D = 4, N = 5 and Q = 1

Main Case 3 :
D = 8
Then N + 6Q = 20-18 = 2
N = 2- 6Q
This forces Q to be 0
N = 2 , Q = 0 , D = 8 , so C = 10
( checking : 1 (10) + 10(8) + 5( 2) + 25(0) = 10+80+10+0 =100 )
Solution :::: c = 10 , D = 8, N = 2 and Q = 0
So, there are 6 possible solutions:
So the following are the six solutions for the problem
Solution 1:::: c = 5 , D = 0, N = 14 and Q = 1
Solution 2:::: c = 10 , D = 0, N = 8 and Q = 2
Solution 3:::: c = 15 , D = 0, N = 2 and Q = 3
Solution 4:::: c = 5 , D = 4, N = 11 and Q = 0
Solution 5:::: c = 10 , D = 4, N = 5 and Q = 1
Solution 6:::: c = 10 , D = 8, N = 2 and Q = 0
I hope this explains how we get the number of coins each to make 1 dollar(100cents/pennies) and also total of 100 coins.(both the given conditions satisfied)