Diane has one stamp of each positive integer value 1 cent through 9 cents, inclusive. She wants to put 10 cents worth of postage in a row across the top of an envelope. If arrangements of the same stamps in different order are considered different, how many arrangements are possible?
Make a list of combinations.
2 stamps: 1,9
number of ways to arrange = 2
3 stamps: 1,2,7 - 1,3,6 - 1,4,5 - 2,3,5
number of ways to arrange = 4(3!) = 24
4 stamps: 1,2,3,4 - that is all of them
number of ways to arrange = 1(4!) = 24
total number of ways = 2+24+24 = 50
Try doing it with casework.
Case: 1
1,9 x2
2,8 x2
3,7 x2
4,6 x2
2+2+2+2=8.
That's it for case one since Diane only has one stamp for each.
Case: 2
1,2,7 x6
1,3,6 x6
1,4,5 x6
2,3,5 x6
6+6+6+6=24.
Case: 3
1,2,3,4 x24
The total answer is 8+24+24=56.
The answer is 56
Its 56 u dummy
To find the number of possible arrangements, we can use combinatorics principles.
Since Diane has stamps from 1 cent through 9 cents, we need to find the number of ways she can select stamps with a total value of 10 cents.
To simplify the problem, we can break it down into smaller cases based on the value of the last stamp used.
Case 1: The last stamp used is 1 cent.
In this case, Diane needs to find a combination of stamps adding up to 10 - 1 = 9 cents, which she can do with the remaining stamps. Since the 1 cent stamp is fixed, she has 8 remaining stamps to choose from. We can use combinations to determine the number of ways she can select stamps:
C(8, 8) = 1 way.
Here C(8, 8) represents "8 choose 8" or the number of ways to select all 8 remaining stamps.
Case 2: The last stamp used is 2 cents.
Diane needs to find a combination of stamps adding up to 10 - 2 = 8 cents. Again, she has 8 remaining stamps to choose from. We can use combinations to determine the number of ways she can select stamps:
C(8, 7) = 8 ways.
Here C(8, 7) represents "8 choose 7" or the number of ways to select 7 remaining stamps.
We can continue this process for each possible value of the last stamp, and then sum up the results to get the total number of arrangements.
Case 1: C(8, 8)
Case 2: C(8, 7)
Case 3: C(8, 6)
Case 4: C(8, 5)
Case 5: C(8, 4)
Case 6: C(8, 3)
Case 7: C(8, 2)
Case 8: C(8, 1)
Case 9: C(8, 0)
Adding these together, we get:
C(8, 8) + C(8, 7) + C(8, 6) + C(8, 5) + C(8, 4) + C(8, 3) + C(8, 2) + C(8, 1) + C(8, 0)
Using the formula for combinations, C(n, r) = n! / (r!(n-r)!), we can calculate each of these terms:
C(8, 8) = 8! / (8!(8-8)!) = 1
C(8, 7) = 8! / (7!(8-7)!) = 8
C(8, 6) = 8! / (6!(8-6)!) = 28
C(8, 5) = 8! / (5!(8-5)!) = 56
C(8, 4) = 8! / (4!(8-4)!) = 70
C(8, 3) = 8! / (3!(8-3)!) = 56
C(8, 2) = 8! / (2!(8-2)!) = 28
C(8, 1) = 8! / (1!(8-1)!) = 8
C(8, 0) = 8! / (0!(8-0)!) = 1
Now we can sum up these terms:
1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256
Therefore, there are 256 possible arrangements for Diane to put 10 cents worth of postage in a row across the top of an envelope.