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in the 2nd question it is said that 3-digit area code can not begin with 0 or 1, so it means 7-digit local can begin with them. However, you did not use 0 and 1 for 7-digit. Please correct me if I am mistaken. Thanks in advance for everything.
Based on how the second question is asked, shouldn't it only use the probability of that one single-starting area code of its 8 possible outcomes instead of including the first digit within the middle/second column of a telephone number? I'm not sure if I'm reading the question correctly or not.
For the thrid problem the answer is 100. Since this is a multiple choice quiz, we have 25 ways to answer a question (let's say our options are A, B, C, D and E): A B C D E AB BA CA DA EA ABC BAC CAB DAB EAB ABCD BACD CABD DABC EABC ABCDE BACDE CABDE DABCE EABCD Then, we multiply number of options to anwser one question by 4 because we have 4 questions: 25*4=100
Not exactly. Because of the context of the problem, you solve it differently than the first problem. In #1, you don't repeat any letters because it wouldn't make sense to wear 2 pairs of pants at the same time. However, in a multiple choice test you could pick A for all four problems.
There's actually 31 ways to answer a question since 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 31. If you include no answers, then it's 32. In a multiple choice question, you don't use permutation (nPr), because you don't need to answer them in order. ABCDE, BACDE, CABDE, DABCE and EABCD are the same. Instead, use combination (nCr). Anyway, neither permutation nor combination works here. Idk why tho. My brain just can't anymore.
Wait about that second example sir If the first digit of the seven digit code isn't going to move from that position wouldn't it be like (8*10²) + (8*10⁸)
I don't really get your question but hope this explanation will help you. The fundamental counting principle states if there is n way to do smth and m ways to do another things, the total possible outcomes would be m*n. So there is a limitation in the first 3 digits and local number (not 0 or 1), you can only choose from 2 to 9, which is only 8 outcomes. For the rest, you can choose from 10. And if you want to find total possibilities, multiply everything.
@@AliceKMay just a quick question since we already used the number 8, wouldn't it become 7 for the beginning of the 7-digit number after the 3 area code. because 7 isn't 1 or 0 is there any reason why. I hope this is making sense and that reply was awesome great explanation there.
@@abdulhamidsiad7846 I think you're confused with digits and possibilities. In this problem, we are thinking how many digits are possible to be added in a telephone number. 8 means 8 possible outcomes (2, 3, 4, 5, 6, 7, 8, 9). Not the digit 8. Hope this helps!
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Just a note, for the tree diagram you can also count the last row for the number of possibilities instead of counting each and every possibility.
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in the 2nd question it is said that 3-digit area code can not begin with 0 or 1, so it means 7-digit local can begin with them. However, you did not use 0 and 1 for 7-digit. Please correct me if I am mistaken. Thanks in advance for everything.
In the question, both the 3-digit area code and 7-digit local telephone number cannot begin with 0 or 1
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for the second question, can we use the permutation to get the answer? Always confused about the difference and when should I use these ways
Based on how the second question is asked, shouldn't it only use the probability of that one single-starting area code of its 8 possible outcomes instead of including the first digit within the middle/second column of a telephone number?
I'm not sure if I'm reading the question correctly or not.
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For the thrid problem the answer is 100.
Since this is a multiple choice quiz, we have 25 ways to answer a question
(let's say our options are A, B, C, D and E):
A B C D E
AB BA CA DA EA
ABC BAC CAB DAB EAB
ABCD BACD CABD DABC EABC
ABCDE BACDE CABDE DABCE EABCD
Then, we multiply number of options to anwser one question by 4 because we have 4 questions: 25*4=100
Not exactly. Because of the context of the problem, you solve it differently than the first problem. In #1, you don't repeat any letters because it wouldn't make sense to wear 2 pairs of pants at the same time. However, in a multiple choice test you could pick A for all four problems.
There's actually 31 ways to answer a question since 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 31. If you include no answers, then it's 32. In a multiple choice question, you don't use permutation (nPr), because you don't need to answer them in order. ABCDE, BACDE, CABDE, DABCE and EABCD are the same. Instead, use combination (nCr).
Anyway, neither permutation nor combination works here. Idk why tho. My brain just can't anymore.
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Wait about that second example sir
If the first digit of the seven digit code isn't going to move from that position wouldn't it be like
(8*10²) + (8*10⁸)
I don't really get your question but hope this explanation will help you.
The fundamental counting principle states if there is n way to do smth and m ways to do another things, the total possible outcomes would be m*n.
So there is a limitation in the first 3 digits and local number (not 0 or 1), you can only choose from 2 to 9, which is only 8 outcomes. For the rest, you can choose from 10.
And if you want to find total possibilities, multiply everything.
@@AliceKMay just a quick question since we already used the number 8, wouldn't it become 7 for the beginning of the 7-digit number after the 3 area code. because 7 isn't 1 or 0 is there any reason why. I hope this is making sense and that reply was awesome great explanation there.
@@abdulhamidsiad7846 I think you're confused with digits and possibilities. In this problem, we are thinking how many digits are possible to be added in a telephone number.
8 means 8 possible outcomes (2, 3, 4, 5, 6, 7, 8, 9). Not the digit 8.
Hope this helps!
@@AliceKMay thanks, cleared my doubt
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Can this work with solving a rubix cube?
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The only thing I am wondering is why Mike has such a weird taste in fashion 😂
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