I dont get my mistake ; combinatory analysis

[Marichou]

Hello, I know my answer is wrong but I can't tell what I am doing wrong

How many possibilities are there for a 4-digit code that must either start with 0 or end with 2.


My answer

If it starts with 0:

1*10*10*9

If it ends with 2:

9*10* 10*1

Addition of the 2 possibilities: 900+900
That is 1800 possibilities.


The correct answer

If it ends with 2:

10*10*10*1

If it starts with 0:

1*10*10*10

Code which ends with 2 and starts with 0:

1*10*10*1

Addition of 2 possibilities minus the possibilities where the code ends with 2 and starts 0:

1000+1000-100 That is 1900 possibilities

 

[stapel]

Hello, I know my answer is wrong but I can't tell what I am doing wrong

How many possibilities are there for a 4-digit code that must either start with 0 or end with 2.

My answer

If it starts with 0:

1*10*10*9

Why are you saying that there are only nine options for the fourth digit?

If it ends with 2:

9*10* 10*1

Why do you say that there are only nine options for the first digit? What is your reasoning?

 

[Dr.Peterson]

If it starts with 0:

1*10*10*9

If it ends with 2:

9*10* 10*1

Addition of the 2 possibilities: 900+900
That is 1800 possibilities.

It looks like your first case is "starts with 0, doesn't end with 2", and the second is "ends with 2, doesn't start with 0".

What about if it starts with 0 and ends with 2??

You can either add in that case, or include it in one of the others. (Or you could use the rule for "or".)

 

[pka]

Hello, I know my answer is wrong but I can't tell what I am doing wrong
How many possibilities are there for a 4-digit code that must either start with 0 or end with 2.
The correct answer
If it ends with 2:
10*10*10*1
If it starts with 0:
1*10*10*10
Code which ends with 2 and starts with 0:
1*10*10*1
Addition of 2 possibilities minus the possibilities where the code ends with 2 and starts 0:
1000+1000-100 That is 1900 possibilities

I agree with the given answer. But here is the basic counting rule.
If [imath]\mathcal{A}[/imath] is the number that begin with [imath]0[/imath] & [imath]\mathcal{B}[/imath] is the number that end with [imath]2[/imath]
There are [imath]1\cdot 10^3=1000[/imath] codes that begin with a [imath]\bf{0}[/imath]
There are [imath]1\cdot 10^3=1000[/imath] codes that end with a [imath]\bf{2}[/imath]
There are [imath]1\cdot 10^2=100[/imath] codes that begin with a [imath]\bf{0}\text{ and end with a }\bf{2}[/imath]
The number which either begin with a [imath]\bf{0}[/imath] or end with [imath]\bf{2}[/imath] is:
[imath]\#(\mathcal{A}\cup\mathcal{B})=\#(\mathcal{A})+\#(\mathcal{B})-\#(\mathcal{A}\cap\mathcal{B})=1900[/imath][imath][/imath][imath][/imath]

 

[Marichou]

Why are you saying that there are only nine options for the fourth digit?



Why do you say that there are only nine options for the first digit? What is your reasoning?

Because it can't start with 0 and can't end with 2

 

[Marichou]

It looks like your first case is "starts with 0, doesn't end with 2", and the second is "ends with 2, doesn't start with 0".

What about if it starts with 0 and ends with 2??

You can either add in that case, or include it in one of the others. (Or you could use the rule for "or".)

Because it can't be the two at same time (starts with 0 ends 2).

The difference between my response and the correct one is :

I calculated the codes
"starting with 0 and NOT ending with 2"
and (+)
"Notstarting with 0 and ending with 2"

And the other

Calculated all the response possible minus "starts with 0 ends with 2"
Because it cant have the two condition at the same time.


I'll try asking a teacher on friday.

 

[stapel]

How many possibilities are there for a 4-digit code that must either start with 0 or end with 2.

Why are you saying that there are only nine options for the fourth digit?
Why do you say that there are only nine options for the first digit? What is your reasoning?

Because it can't start with 0 and can't end with 2

Actually, the "either" in the exercise statement is important. It says that a code can EITHER start with zero (with no conditions on the rest of the digits) OR ELSE end with 2 (with no conditions on the rest of the digits). It does not say that the code cannot start with zero AND cannot end with 2.

I think you're trying to answer the "AND" version, but the correct solution is for the "EITHER/OR ELSE" version.

 

[khansaheb]

Because it can't be the two at same time (starts with 0 ends 2).

The difference between my response and the correct one is :

I calculated the codes
"starting with 0 and NOT ending with 2"
and (+)
"Notstarting with 0 and ending with 2"

And the other

Calculated all the response possible minus "starts with 0 ends with 2"
Because it cant have the two condition at the same time.


I'll try asking a teacher on friday.

Read the response #4 & #7 carefully. That will explain to you the thought process required for this problem. How do the thought processes differ in #4 vs. #7?