Please help with this problem.

In a student writing contest with 25 entries, 3 essays are selected for first, second, and third place awards and 5 are selected for honorable mention. How many ways can this be done? Thanks.

Ok Gen, for the first place there are 25 choices, for second place 24 choices and for third place 23. There are thus 25-times-24-times-23 permutations for the first three places
Would you agree and do you see why?
That leaves 22 remaining from which we can select 5 in any order. That means there are 22 choose 5 combinations for honorable mention.
The total ways this can be done is the product of these two numbers:
(permutations of 25 with 3) times (combinations of 22 choose 5).
Do you see why?
In general, when order matters think permutations. When it doesn't, think combinations.

Thank you Roger. I do understand the order of the first 3, and the non order of the 5. I just did not know that permutations and combinations could be combined. When I do this I get 363,409,200. This is a huge number.Is this correct?

That looks correct by my calculations.
Both permutations and combinations use factorials and these grow very quickly. So numbers this size are common.

Thank you very much Roger.

Find an equation of the line containing the given pair of points (3,5)and (9,8)

assume that there are 4 dimes, 3 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?

Explain Bot Explain Bot answered
9 months ago

To find the equation of the line containing the given pair of points (3,5) and (9,8), you can use the point-slope formula. First, calculate the slope (m) using the formula (y2 - y1) / (x2 - x1) where (x1, y1) = (3,5) and (x2, y2) = (9,8).

m = (8 - 5) / (9 - 3)
m = 3 / 6
m = 1/2

Now that you have the slope, you can use the point-slope formula y - y1 = m(x - x1), where (x1, y1) is any of the given points. Let's use (3,5) as (x1, y1) in this case.

y - 5 = (1/2)(x - 3)

This is the equation of the line containing the points (3,5) and (9,8). You can simplify it further if needed.

Now, let's move on to the next question.

To find the number of possible ways the selection of coins can be made so that the value is at least 25 cents, we can break it down into cases.

Case 1: Using only quarters
Since the value should be at least 25 cents, we can select 1 or 2 quarters.

If we select 1 quarter, then we need to select the remaining amount using dimes and nickels. So there are 4 ways to select 1 quarter and the remaining value using dimes and nickels.

If we select 2 quarters, then we don't need to select any more coins. So there is 1 way to select 2 quarters without any other coins.

Total ways with only quarters = 4 + 1 = 5

Case 2: Using at least one nickel
If we use at least one nickel, then we can select the remaining amount using dimes, nickels, and quarters.

If we use 1 nickel, then the remaining amount should be at least 20 cents. So we can select dimes and/or quarters to make up the remaining value. There are 4 ways to select 1 nickel and the remaining value using dimes and quarters.

If we use 2 nickels, then the remaining amount should be at least 15 cents. Again, we can select dimes and/or quarters to make up the remaining value. There are 3 ways to select 2 nickels and the remaining value using dimes and quarters.

If we use 3 nickels, then the remaining amount should be at least 10 cents. We can select dimes and/or quarters to make up the remaining value. There are 2 ways to select 3 nickels and the remaining value using dimes and quarters.

If we use 4 nickels, then the remaining amount should be at least 5 cents. We can select dimes and/or quarters to make up the remaining value. There is 1 way to select 4 nickels and the remaining value using dimes and quarters.

Total ways with at least one nickel = 4 + 3 + 2 + 1 = 10

Case 3: Using only dimes and nickels
If we don't use any quarters, then we can only use dimes and/or nickels.

If we use 1 dime, then the remaining amount should be at least 15 cents. We can select nickels to make up the remaining value. There are 3 ways to select 1 dime and the remaining value using nickels.

If we use 2 dimes, then the remaining amount should be at least 5 cents. We can select nickels to make up the remaining value. There is 1 way to select 2 dimes and the remaining value using nickels.

If we use 3 dimes, then the remaining amount should be at least 5 cents. Again, we can select nickels to make up the remaining value. There is 1 way to select 3 dimes and the remaining value using nickels.

Total ways with only dimes and nickels = 3 + 1 + 1 = 5

Total ways = Total ways with only quarters + Total ways with at least one nickel + Total ways with only dimes and nickels
Total ways = 5 + 10 + 5 = 20

Therefore, there are 20 possible ways to make the selection of coins so that the value is at least 25 cents.