You have unlimited stocks of 50p, £1 and £2 coins. How many ways (combinations - order does not matter) are there of making up £100 with these coins?
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It's easy to formulate an equation. Let the number of coins 50p, £1 and £2 are x , y , a n d z respectively.
Thus:
0 . 5 x + y + 2 z = 1 0 0
x + 2 y + 4 z = 2 0 0 ........................................ 1
Since 2 y a n d 4 z are always even, x must be also even number.
Let:
x = 2 k .............................................. 2
x in 1 substituted with 2
x + 2 y + 4 z = 2 0 0 ⇒ 2 k + 2 y + 4 z = 2 0 0
k + y + 2 z = 1 0 0 .............................................. 3
Since 2 z is always even, then easily find the number of making up to 1 0 0
k + y + 2 z = 1 0 0
k + y = 1 0 0 − 2 z
The value of 1 0 0 − 2 z must be non-negative integer and z must be non-negative integer. It is easy that the boundaries of z are 0 ≤ z ≤ 5 0
The number of choosing k and y is calculated by using combinations with repetition, since we know z has the bounds, we conclude that
z = 0 ∑ 5 0 ( 1 0 0 − 2 z 1 0 0 − 2 z + 1 )
= z = 0 ∑ 5 0 ( 1 0 0 − 2 z 1 0 1 − 2 z )
= z = 0 ∑ 5 0 ( 1 1 0 1 − 2 z )
= 1 0 1 + 9 9 + 9 7 + . . . + 3 + 1 = ( 2 1 0 1 + 1 ) 2 = 5 1 2 = 2 6 0 1