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Sagot :
Problem 8, part (i)
Since repetition is not allowed, we cannot use the digits from this set: {5,6,2}
So the digits we can use are: {0,1,3,4,7,8,9} which consists of 10-3 = 7 digits.
We have 7 choices for the fourth slot, 6 choices for the fifth slot, and 5 choices for the last slot. That gives 7*6*5 = 42*5 = 210 permutations of 3 digits. So there are 210 different phone numbers of the form 562-ABC, where A,B,C are digits selected from the set {0,1,3,4,7,8,9} and repeats are not allowed.
An example phone number could be 562-789
Answer: 210
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Problem 8, part (ii)
Repeats are allowed so we can pick anything from {0,1,2,3,4,5,6,7,8,9} which is a set of 10 items.
We have 10 choices for slot four, 10 choices for slot five, and 10 choices for slot six. We get 10*10*10 = 1000 different three-digit sequences. So there are 1000 different phone numbers of the form 562-ABC where A,B,C are digits from {0,1,2,3,4,5,6,7,8,9} and repeats are allowed.
An example phone number could be 562-555
Answer: 1000
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Problem 9
We're not allowed to have 0 in the first slot, but we can have it in any of the two remaining slots. We're only allowed to have exactly one copy of zero.
If 0 is in the second slot, then there are 9*9 = 81 different phone numbers. The same can be said if 0 is in the third slot. Therefore, we have 2*81 = 162 different phone numbers in this format. Some example three digit phone numbers are: 102, 510, 670.
Answer: 162
Answer:
#8
The 6-digit number is 562 xyz.
x, y, z can be any digit from 0 to 9, total 10 options.
(i) No repetition
For x- we can have 10 - 3 = 7 options (except 5, 6, 2)
For y- we can have 10 - 3 - 1 = 6 options (except 5, 6, 2, x)
For z- we can have 10 - 3 - 2 = 5 options (except 5, 6, 2, x, y)
Number of options:
- 7*6*5 = 210
(ii) Repetition allowed
x, y, z can have 10 options each.
Number of options:
- 10*10*10 = 1000
#9
The 3 digit number is xyz.
- x- have 9 options (except zero)
- If y = 0, there are 9 options for z
- If z = 0, there are 9 options for y
Number of options:
- 9*9 + 9*9 = 162
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