Teo has a combination of quarters, loonies, and toonies in his pocket. He knows he has 9 coins in total and that he has at least one of each
type of coin.
a) What is the probability that Teo has more than $8?

b) How many different combinations of coins could Teo use to pay for an item
that costs between $10 and $12, if he uses fewer toonies than loonies?

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Step-by-step explanation:

343

Step-by-step explanation:

There are quarters(25¢), loonies ($1), and toonies ($2)

And total number of coins is 9

If loonies ($1), and toonies ($2) are one, then quarters(25¢) is 7

If quarters(25¢), and toonies ($2) are one, then loonies ($1)is 7

If loonies ($1), and quarters(25¢) are one, then toonies ($2) is 7

So possible number number of combination for an item that cost between $10 and $12 is

= 1^3 * 1^3 * 7^3

= 343

is should be y = 5x

step-by-step explanation:

correct me if im wrong