At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
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
=============================================================
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
=============================================================
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
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.