How many sequences of four numbers exist with these conditions?

Question

First of all, this may seem very basic for you, but I've never been good at math, and I can't figure this out!

I need to solve the number of sequences that exist under this conditions:

• All sequences are made of four numbers like 0123
• All the numbers must be different
• The first and the last number are on the same horizontal line
• The second and third numbers are on the horizontal line immediately above the first and last number line

This is table that is provided with the problem

[1][2][3]
[4][5][6] 
[7][8][9] 
[-][0][+]

How many sequences exist under these conditions?

a) 72

b) 90

c) 162

d) 171

Answer 1

As two pairs of different digits must be on the same row, $0$ is excluded.

You can choose the first and last digits freely in the second or third row, which makes $2\cdot3\cdot2$ options, then the second/third digits in the first or second row respectively.

In total,

$$2\cdot3\cdot2\cdot3\cdot2.$$

Answer 2

You shall only choose the first and last digit from the two middle rows. That's because the bottom row contains only one number and for the top row, there's no row above to choose the second and third digit.

Let's start by choosing two digits from the $3$rd row. We can choose and rearrange two items out of three in $$P(3,2)=\frac{3!}{1!}=6$$ ways. Following the exact same reasoning, for the second and third digit, we have: $$ P(3,2)=\frac{3!}{1!}=6 $$ ways for the middle digits.

Thus, $6$ ways to choose first and last digits and $6$ ways to choose second and third. That's a total of $36$ ways to choose all $4$ digits.

Also, we could repeat all these actions, choosing the first and last from the $2$nd row and the middle ones from the $1$st one. Again, that's a total of $36$ ways to choose all four digits.

At last, since we can carry out this procedure either the first $36$ ways or the other $36$ ways, the total ways are $$ 36+36=72. $$