How many different sums of money are possible using at least 3 coins from a collection consisting of one penny, one nickel, one dime, on quarter and one loonie?
There are 5C3+5C4+5C5 ways to choose the coins.
Now, are there any sums which can be made in more than one way? If so, then they'd be counted twice.
The most cases come from 3 coins at a time, 5C3 as Steve stated, there are 10 of those.
Let's consider any duplication of sums in those.
PND -- 16 cents
PNQ -- 31
PNL -- 106
PDQ -- 36
PDL -- 111
PQL -- 126
NDQ -- 40
NDL -- 115
NQL -- 130
DQL -- 135
Four at a time:
PNDQ - 41
PNDL -- 116
PNQL -- 131
PDQL -- 136
NDQL -- 140
And lastly, take all 5
PNDQL --- 141
Fortunately, there are no duplication of sums, so the total number of choices is 16
check my addition in the sums.
To find the number of different sums of money using at least 3 coins from the given collection, we can use a systematic approach.
First, let's start by listing all the different individual coin values in cents:
- Penny: 1 cent
- Nickel: 5 cents
- Dime: 10 cents
- Quarter: 25 cents
- Loonie: 100 cents
Now, let's consider the possible scenarios when choosing at least 3 coins. We have a total of 5 coins, so we can make the following combinations:
1) Choosing 3 coins:
We have 5C3 combinations, which is equal to (5!)/(3!(5-3)!), where "!" denotes the factorial.
5C3 = (5!)/(3!2!) = 10 combinations
2) Choosing 4 coins:
We have 5C4 combinations, which is equal to (5!)/(4!(5-4)!) = 5 combinations.
3) Choosing all 5 coins:
We have 5C5 combinations, which is equal to 1 combination.
Now, let's sum up the combinations:
Number of different sums of money = (Number of combinations when choosing 3 coins) + (Number of combinations when choosing 4 coins) + (Number of combinations when choosing all 5 coins)
= 10 combinations + 5 combinations + 1 combination
= 16 different sums of money.
Therefore, there are a total of 16 different sums of money possible when using at least 3 coins from the given collection.