Finding the number of $4$-digit numbers so no two adjacent digits are even

Question

Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.

Answer 1

You can approach this thinking that you have two different subsets: the even numbers $\{0,2,4\}$ and the odd numbers $\{1,3,5\}$. In order to avoid adjacent even numbers you have the two following cases:

Alternating even and odd numbers, starting with an odd number:

$3 \cdot 3 \cdot 2 \cdot 2 = 36$ possible combinations

First you can place any of the digits from $\{1,3,5\}$, then any from $\{0,2,4\}$, in the third place any of the remaining digits of the odd numbers set (2 in total), and same case for the even numbers in the last digit.

Alternating even and odd numbers, starting with an even number:

$2 \cdot 3 \cdot 2 \cdot 2 = 24$ possible combinations

The trick here is that starting with $0$ doesn't lead to a 4 digit number (you can add the zeros you want to the left of a number and it remains the same number).

For even numbers at both extremes:

$2 \cdot 3 \cdot 2 \cdot 2 = 24$

For only one even number:

$2 \cdot 3 \cdot 2 \cdot 1 = 12$

$3 \cdot 3 \cdot 2 \cdot 1 = 18$

$3 \cdot 2 \cdot 3 \cdot 1 = 18$

$3 \cdot 2 \cdot 1 \cdot 3 = 18$

TOTAL:

Therefore, in total, there are $(24+36+18\cdot 3+12+24=150)$ possibilities

Answer 2

Even digits $\{0,2,4\}$

Odd digits $\{1,3,4\}$

Numbers of the format: Even Odd Even Odd

$2 \times 3 \times 2 \times 2=24$

Numbers of the format: **Odd Even Odd Even

$3 \times 3 \times 2 \times 2=36$

Numbers of the format: Odd Odd Odd Even

$3 \times 2 \times 1 \times 3=18$

Numbers of the format: Odd Odd Even Odd

$18$

Numbers of the format: Odd Even Odd Odd

$18$

Numbers of the format: Even Odd Odd Odd

$2 \times 3 \times 2 \times 1=12$

**Numbers of the format: Even Odd Odd Even **

$2 \times 3 \times 2 \times 2=24$

Required count:

$24+36+18+18+18+12+24=150$

Answer 3

Since you have three odd and three even digits, you must use at least one even digit. But you cannot use three even digits as that would force two adjacent even digits.

a) One even digit: There are $3$ ways to choose the even digit, and $3\cdot 2\cdot 1$ ways to choose the odd digits from left to right, and $4$ ways to choose the position of the even one:

$3\cdot 3\cdot 2\cdot 1\cdot 4=72$

b) Two even digits: There are $3\cdot 2$ ways to choose the even digits from left to right, and $3\cdot 2$ ways to choose the odd digits from left to right, and ${4\choose 2}-3$ ways to choose the even positions non-adjacent:

$3\cdot 2\cdot 3\cdot 2\cdot(6-3)=108$

In total, we get

$72+108=180$.

(The above assumes that leading $0$ is allowed.)

Answer 4

Case 1 -

Numbers start with even digit except 0 at first place. Calculate product of number of ways for each digit.

First digit cant be 0. So we have 2 ways to choose first even. Then 3 ways for second digit as odd. Now we can choose 0 also for 3rd digit so 2 ways. 2 ways for last odd digits.

We have 2 × 3 × 2 × 2 = 24 ways.

Case 2 -

Numbers start with odd digit. Calculate product of number of ways for each digit.

3 ways to choose first odd digit. Then 3 ways for second digit as even. For 3rd digit as odd so 2 ways. 2 ways for last even digits.

We have 3 × 3 × 2 × 2 = 36 ways.

Case 3 -

Numbers in brackets for odd numbers.

Only one starting number is even (except 0) and others are odd.

We have 2 × (3) × (2) × (1) = 12

Case 4 -

Numbers in brackets for odd numbers.

Any one of last three digits is even.

We have (3) × 3 × (2) × (1) + (3) × (2) × 3 × (1) + (3) × (2) × (1) × 3 = 54

Case 5 -

Number in brackets for odd numbers.

First and last digit is even and first can't be zero.

2 × (3) × (2) × 2 = 24

Total number of ways = 24 + 36 + 12 + 54 + 24 = 150.