Question 1193916: How many different four-digit numbers are possible if each digit of the number is either 0,1,2,3,4 or 5, and numbers starting with 3 cannot be even?
The textbook answer is 972.
Can someone tell me how to solve this problem?
Found 4 solutions by Edwin McCravy, greenestamps, ikleyn, mccravyedwin:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
First let's calculate how many 4-digit numbers have digits from
{0,1,2,3,4,5},
regardless of whether they start with 3 or are odd or even. Then
we'll calculate how many start with 3 and are even, and subtract.
A. Since a 4-digit number can't start with 0, the ways to
choose the 1st digit are any one of these: {1,2,3,4,5}.
How many choices is that? _____
B. The ways to choose the 2nd digit are any of these:
{0,1,2,3,4,5}. How many choices is that? _____
C. The ways to choose the 3rd digit are any of these:
{0,1,2,3,4,5}. How many choices is that? _____
D. The ways to choose the 4th digit are any of these:
{0,1,2,3,4,5}. How many choices is that? _____
E. Multiply the answers to questions A,B,C, and D together.
What do you get? _____
Now let's calculate how many we need to subtract from the answer
to question E.
We must not include any 4-digit numbers that start with 3 that
are even. So let's see how many there are of those>
F. There is only one choice for the first digit of the numbers
we want to NOT include. How many choices is that? 1
(I filled that in for you as 1 for it's the 1 digit "3").
G. The ways to choose the 2nd digit are any of these:
{0,1,2,3,4,5}. How many choices is that? _____
H. The ways to choose the 3rd digit are any of these:
{0,1,2,3,4,5}. How many choices is that? _____
I. But since the numbers we don't want to count are even, the
last (fourth) digit must be one of these (0, 2, 4). How many
choices is that? _____
J. Multiply the answers for questions F,G,H, and I. What do
you get? _____
K. Subtract the answer to J from the answer to E. What do you
get? _____
Should be 972. If you don't get 972, tell me your answers to the
above questions in the thank-you note form below and I'll get back to
you by email.
Edwin
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The first digit can't be 0; and there are special requirements if the first digit is 3. So consider two cases -- first digit 1, 2, 4, or 5; or first digit 3.
(1) first digit 1, 2, 4, or 5
There are 4 choices for that first digit, then there are no restrictions for the other digits; so the number of 4-digit numbers with first digit 1, 2, 4, or 5 is 4*6*6*6=864.
(2) first digit 3
The first digit has to be 3 (1 choice); the last digit must be odd (1, 3, or 5 -- 3 choices); and there are no restrictions on the other two digits. The number of possible 4-digit even numbers with first digit 3 is 1*3*6*6=108.
The total number of 4-digit numbers with the given requirements is 864+108=972.
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A comment for tutor @ikleyn, whose math is nearly always very good but whose English is often not....
There is nothing deficient in the statement of the problem. It says that "each digit of the number is either 0,1,2,3,4 or 5". That means repetition of digits is allowed.
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
How the problem is worded in the textbook, it is defective formulation, because it NOWHERE says that repetition of digits is not allowed.
Very strange textbook. Probably, it is some home-made feature, which was never peer-reviewed.
Can you post the name of this textbook and its ISBN ?
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Edwin, I see from your post that you try to explain me something. Thanks for your attempt.
But what was written in my post, was not a question "if repeating is allowed or not ?"
It was a statement, saying that the problem formulation is defective, because it does not indicate directly,
if repeating is allowed or not.
When similar problem is formulated, such indication must be presented directly in the problem;
a reader should not extract this info from the answer or to guess it based on the context.
Answer by mccravyedwin(406)  (Show Source):
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