Assume that you have 7 dimes and 5 quarters (all distinct), and you select 4 coins. In how many ways can the selection be made so that at least 3 coins are dimes?In how many ways can the selection be made so that at least 3 coins are dimes?
To find the number of ways to select 4 coins with at least 3 dimes, we need to consider two cases: selecting 3 dimes and 1 quarter, and selecting all 4 dimes.
1. Selecting 3 dimes and 1 quarter:
- There are 7 ways to select 3 dimes from 7 dimes.
- There are 5 ways to select 1 quarter from 5 quarters.
- Therefore, there are 7 * 5 = 35 ways to select 3 dimes and 1 quarter.
2. Selecting all 4 dimes:
- There is only 1 way to select all 4 dimes.
Finally, we add the results from both cases to get the total number of ways to select 4 coins with at least 3 dimes:
35 + 1 = 36.
So, there are 36 ways to make the selection with at least 3 coins being dimes.
To calculate the number of ways to select the coins, we need to consider the two possible scenarios: selecting exactly 3 dimes and selecting all 4 dimes.
1. Selecting exactly 3 dimes:
To select 3 dimes, we have 7 dimes to choose from, so we choose 3 out of 7 dimes in C(7, 3) ways.
For the remaining coin, we can choose any of the 5 quarters, so we choose 1 out of 5 quarters in C(5, 1) ways.
Therefore, the total number of ways to select exactly 3 dimes is C(7, 3) * C(5, 1) = (7 choose 3) * (5 choose 1) = 35 * 5 = 175 ways.
2. Selecting all 4 dimes:
If we want to select all 4 dimes, we simply choose 4 out of the 7 dimes in C(7, 4) ways.
Since in this scenario, we don't select any quarters, we choose 0 out of 5 quarters in C(5, 0) way (which is equal to 1).
Therefore, the total number of ways to select all 4 dimes is C(7, 4) * C(5, 0) = (7 choose 4) * (5 choose 0) = 35 * 1 = 35 ways.
Now, to find the total number of ways to select the coins such that at least 3 are dimes, we sum up the two scenarios:
Total ways = Number of ways selecting exactly 3 dimes + Number of ways selecting all 4 dimes
Total ways = 175 + 35 = 210 ways.
Therefore, there are 210 ways to select 4 coins such that at least 3 of them are dimes.
I see 2 cases
1. 3 coins are dimes, the 4th is a quarter
2. all 4 coins are dimes
number of selections for case 1 : C(7,3)*C(5,1)
= 35*5 = 175
number of selection for case 2: C(7,4) = 35
number of "at least 3 dimes" is 175+35 = 210
btw, did you mean to ask the same question twice?