selecting three (3) coins, but assume that there are 5 dimes, 4 nickels, and 2 quarters.
In how many possible ways can the selection be made so that the value of the coins is at least 25 cents?
To find the number of possible ways to select three coins such that the value is at least 25 cents, we first need to identify the minimum coin values that can contribute to a sum of 25 or more cents.
Since we have 5 dimes (10 cents each) and 2 quarters (25 cents each), the minimum sum we can achieve without any nickels is:
2 quarters (25 cents each) + 1 dime (10 cents each) = 60 cents.
Now, let's consider the contribution of the nickels. Each nickel (5 cents each) can increase the value of a selection by 5 cents. If we have one or more nickels, they can potentially help us reach the minimum target sum of 25 cents.
Let's analyze the possible scenarios:
1. If we select both quarters (50 cents), we need at least one additional coin to reach 25 cents. This can be achieved by selecting one dime (10 cents) and one nickel (5 cents). So, the number of ways to select coins in this scenario is 5 (since we have 5 dimes).
2. If we select one quarter (25 cents) and one additional coin (since we need 25 cents in total), it can be a dime or a nickel.
a. If we select one quarter (25 cents) and one dime (10 cents), we need an additional 15 cents to reach 25 cents. This can be achieved by selecting either one dime (10 cents) and one nickel (5 cents), or three nickels (15 cents). In either case, the number of ways to select coins in this scenario is again 5.
b. If we select one quarter (25 cents) and one nickel (5 cents), we need an additional 20 cents to reach 25 cents. This can be achieved by selecting either two dimes (20 cents), or one dime (10 cents) and two nickels (10 cents). In either case, the number of ways to select coins in this scenario is 4 (since we have 4 nickels).
Therefore, the total number of ways to select three coins such that the value is at least 25 cents is:
Number of ways in Scenario 1 + Number of ways in Scenario 2 = 5 + 5 + 4 = 14.
So, there are 14 possible ways to make the selection.
To calculate the number of possible ways to select three coins such that the value is at least 25 cents, we can analyze the possible coin combinations based on their values.
1) Selecting all three quarters: There are 2 quarters, so there is only 1 way to choose 3 quarters.
2) Selecting two quarters and one dime: There are 2 quarters and 5 dimes. We can choose 2 quarters in 2 ways and 1 dime in 5 ways. So, there are 2 * 5 = 10 ways to select two quarters and one dime.
3) Selecting one quarter, one dime, and one nickel: There are 2 quarters, 5 dimes, and 4 nickels. We can choose 1 quarter in 2 ways, 1 dime in 5 ways, and 1 nickel in 4 ways. So, there are 2 * 5 * 4 = 40 ways to select one quarter, one dime, and one nickel.
4) Selecting one quarter, one dime, and two nickels: There are 2 quarters, 5 dimes, and 4 nickels. We can choose 1 quarter in 2 ways, 1 dime in 5 ways, and 2 nickels in (4 choose 2) = 6 ways. So, there are 2 * 5 * 6 = 60 ways to select one quarter, one dime, and two nickels.
Adding up the possibilities:
1 + 10 + 40 + 60 = 111
Therefore, there are 111 possible ways to select three coins such that the value is at least 25 cents.
The only possible cases are, hope I did not miss any
DDD - 30
DDN - 25
DDQ - 45
DNN - 20
DNQ - 40
DQQ - 60
NNN - 15
NNQ - 35
NQQ - 55
How many of these amount to at least 25 cents?
(notice that the fact that we have 5 dimes and 4 nickels is of no consequence, since we can use at most 3, but the 2 quarters limits the cases)