you have eight $1coins and five $2 coins in your pocket. What is the smallest number of coins you need to take out of your pocket to be certain of having atleast one of each coin?

worst case: all 8 $1 coins drawn first.

Then the 9th coin must be a $2 coin.

I fan the help

To be certain of having at least one of each coin, you will need to take out a maximum number of coins.

You can start by taking out all the coins of the same denomination.

Let's take out all the $2 coins first since there are fewer of them.

So, you will need to take out 5 coins (all the $2 coins) to be certain of having at least one $2 coin.

Now, you need to ensure that you have at least one $1 coin.

Since you have 8 $1 coins, you can take out 7 of them and still be left with at least one $1 coin.

Therefore, the smallest number of coins you need to take out of your pocket to be certain of having at least one of each coin is 5 + 7 = 12 coins.

To be certain of having at least one of each coin, you would need to take out a maximum number of coins. In this scenario, there are two possibilities: either all the coins you take out are $1 coins, or all the coins you take out are $2 coins.

In the first case, if you take out all eight $1 coins, you would still not have a $2 coin. In the second case, if you take out all five $2 coins, you would still not have a $1 coin.

Therefore, to be certain of having at least one of each coin, you would need to take out a maximum of 8 + 5 = 13 coins.