[Math] The digits 0-7 are used to form 4-digit numbers…

combinatorics

Restrictions for all questions: No repetition and the thousands place cannot be 0 (ex. 0123, 0555, etc.)

A. How many 4-digit numbers are possible?

B. How many of these 4-digit numbers are odd?

C. How many of these 4-digit numbers are even?

D. How many of these 4-digit numbers are greater than 3100?

E. How many of these 4-digit numbers are less than 3100?

My Answers:

A. 1680, 1470 or 1344.

B 840, 735, or 672

C. Same as B because $B + C = A$

D. 900 (listed down)

E. 450 (listed down)

The main problem of this equation is that we're not allowed to have 0 as our thousands, which makes a seemingly huge contradiction to… everything! My friend and I listed it all down and we're not even sure if it's right or not.
If we work the digits left to right, we get an answer, but if we work right to left, we get a totally different answer!

Is there even a possible formula/equation for this or is this question a lost cause and we really have to list down everything?

Best Answer

How many of these four-digit numbers can be formed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ if no two digits are the same?

Since $0$ cannot be placed in the thousands place, we have seven choices for the thousands digit, which leaves seven choices for the hundreds digit (since we can now use $0$ but cannot select the number we used for the thousands digit), six choices for the tens digit (since we cannot use our choices for the thousands digit or hundreds digit), and five choices for the units digit (since we cannot use our choices for the thousands digit, hundreds digit, or tens digit). Hence, there are $7 \cdot 7 \cdot 6 \cdot 5 = 1470$ such numbers.

Of these four-digit numbers, how many are odd?

We must first impose the restriction that the number is odd, which leaves us with four choices for the units digit ($1$, $3$, $5$, or $7$). Doing so eliminates one of the seven permissible choices for the thousands digit, leaving six choices for the thousands digit. We are now left with six choices for the tens digit and five choices for the units digit. Hence, there are $4 \cdot 6 \cdot 6 \cdot 5 = 720$ odd numbers.

How many of the four-digit numbers are even?

Subtract the number of odd numbers from the total.

Edit: We could also make a direct calculation. We consider two cases, depending on whether or not the last digit is zero.

  1. The last digit is zero. We have one choice for the units digit. Since any of the other digits may be used for the thousands digit, we have seven choices for the thousands digit. This leaves us with six choices for the hundreds digit and five choices for the tens digit, so there are $7 \cdot 6 \cdot 5 \cdot 1 = 210$ such numbers.
  2. The last digit is not zero. Since the number is even, this leaves us with three choices for the units digit ($2$, $4$, or $6$). Since we cannot use either $0$ or the units digit in the thousands place, we are left with six choices for the thousands digit. This leaves us with six choices for the hundreds digit and five choices for the tens digit, so there are $6 \cdot 6 \cdot 5 \cdot 3 = 540$ such numbers.

Since the two cases are mutually exclusive, there are a total of $210 + 540 = 750$ four-digit even numbers.

How many of the four-digit numbers are greater than $3100$?

Notice that this condition imposes a restriction on the thousands digit. If the thousands digit is $3$, it imposes a restriction on the hundreds digit. If the thousands digit is $3$ and the hundreds digit is $1$, there are no further restrictions since it is not possible to obtain $3100$ because no digit may be used more than once.

We consider cases:

  1. The number is at least $4000$. We have four choices for the thousands digit ($4$, $5$, $6$, or $7$). This leaves us with seven choices for the hundreds digit, six choices for the tens digit, and five choices for the units digit. Hence, there are $4 \cdot 7 \cdot 6 \cdot 5 = 840$ possible numbers in this case.
  2. The number $n$ satisfies $3200 \leq n < 4000$. We have one choice for the thousands digit. Since $3$ has been used for the thousands digit, this leaves us with five choices for the hundreds digit ($2$, $4$, $5$, $6$, or $7$). Once the hundreds digit has been selected, we are left with six choices for the tens digit and five choices for the units digit. There are $1 \cdot 5 \cdot 6 \cdot 5 = 150$ choices in this case.
  3. The number $n$ satisfies $3100 < n < 3200$. We have choice for the thousands digit and one choice for the hundreds digit. Since $3$ and $1$ cannot be used, we have six choices for the tens digit and five choices for the units digit. Thus, there are $1 \cdot 1 \cdot 6 \cdot 5 = 30$ choices in this case.

Since the three cases are disjoint, there are a total of $840 + 150 + 30 = 1020$ four-digit numbers greater than $3100$.

How many of these four-digit numbers are less than $3100$?

Since $3100$ is not a possible outcome, subtract the number of four-digit number greater than $3100$ from the total.

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