MHB How Many 3-Digit Even Numbers Can Be Formed from Digits 0-6 Without Repetition?

[paulmdrdo1]

How many 3 digit even numbers can be formed from 0, 1, 2, 3, 4, 5 and 6 with no repetition?

My attempt:

$\frac{5}{H} \times \frac{6}{T} \frac{0}{U} = $ 30 numbers ending with zero not including two digit numbers starting with zero$\frac{4}{H} \times \frac{6}{T} \frac{2}{U} = $ 24 numbers ending with two not including two digit numbers starting with zero$\frac{4}{H} \times \frac{6}{T} \frac{4}{U} = $ 24 numbers ending with four not including two digit numbers starting with zero$\frac{4}{H} \times \frac{6}{T} \frac{6}{U} = $ 24 numbers ending with 6 not including two digit numbers starting with zero

$\therefore$ $30+72= 102$ three-digit even numbers. But in the book's answer key it says 105. What went wrong in my solution? Please help.

 

[Greg]

Set of numbers ending in 0: 6 * 5 = 30 (6 choices for the first digit, 5 for the second digit)
Set of numbers ending in 2: 5 * 5 = 25 (5 choices for the first digit, 5 for the second digit)
Set of numbers ending in 4: 5 * 5 = 25 (5 choices for the first digit, 5 for the second digit)
Set of numbers ending in 6: 5 * 5 = 25 (5 choices for the first digit, 5 for the second digit)

Total: 105

 

[paulmdrdo1]

Set of numbers ending in 0: 6 * 5 = 30 (6 choices for the first digit, 5 for the second digit)
Set of numbers ending in 2: 5 * 5 = 25 (5 choices for the first digit, 5 for the second digit)
Set of numbers ending in 4: 5 * 5 = 25 (5 choices for the first digit, 5 for the second digit)
Set of numbers ending in 6: 5 * 5 = 25 (5 choices for the first digit, 5 for the second digit)

Total: 105

Arent we suppose to go from units to hundreds digit?

 

[Greg]

As we're holding the last digit fixed, only the hundreds (first digit in my post above) and tens (second digit in my post above) need to be counted.

First line: we have 6 choices for the first digit (as we omit 0) and 5 choices for the second digit (we omit 0 as it's the last digit and we omit the hundreds digit).

Second line: with 2 as the last digit, we omit 0 and 2 from the choices for the hundreds digit, leaving 5 digits. With the tens digit we omit 2 and whatever the first digit is for a total of 5 choices and so on for the following lines.

Does that clear things up?

Note that order doesn't matter here - we may count in other ways but we'll still end up with a total of 105.

 

[paulmdrdo1]

As we're holding the last digit fixed, only the hundreds (first digit in my post above) and tens (second digit in my post above) need to be counted.

First line: we have 6 choices for the first digit (as we omit 0) and 5 choices for the second digit (we omit 0 as it's the last digit and we omit the hundreds digit).

Second line: with 2 as the last digit, we omit 0 and 2 from the choices for the hundreds digit, leaving 5 digits. With the tens digit we omit 2 and whatever the first digit is for a total of 5 choices and so on for the following lines.

Does that clear things up?

Note that order doesn't matter here - we may count in other ways but we'll still end up with a total of 105.

hello greg!

Can you pin point what I did wrong in my attempt?
what I did in my solution was, after putting 0 in the units digit, I counted the possible choices for tens digit first rather than hundreds digit. Thus giving me 6 choices for the tens place as I omit 0. And 5 choices for hundreds digit omitting zero and number in the tens digit giving 30 set of numbers with 0 as its units digit.

Same thing can be said for the set of numbers ending with 2. Setting 2 as units digit I will have 6 choices for the tens digit(omitting 2) and 4 choices for hundreds digit (omitting 2, 0 and the tens digit) thus giving me 24 set of 3-digit numbers ending with 2. son on and so forth. please bear with me.

 

[Greg]

My apologies - I was incorrect: order does matter.

We should count all 5 possibilities for the hundreds digit when the last digit is greater than 0; your method eliminates a possible leading digit. If you're still confused try making a list of all the possible numbers ending with 2.

 

[mrtwhs]

I think Greg is right with 105.

We have two mutually exclusive cases. Either the last digit is zero or the last digit is not zero.

If the last digit is zero, this leaves 6 choices for first digit and 5 for the second for a total of $$6 \times 5 = 30$$ ways.

If the last digit is not zero, then the last digit must be a 2,4, or 6. So we have 3 ways for last digit. This leaves 5 choices for the first digit (zero is not allowed) and then we can include zero so there are 5 choices for second digit. So this gives us $$3 \times 5 \times 5 = 75$$ ways.

Thus the total is $$30 + 75 = 105$$ ways.

 

[Wilmer]

Ye olde "long way":

1) (Using 7 digits) 3digit numbers, all digits different: 7! / (7-3)! = 210

2) Remove the 30 cases with 0 as leading digit: 210 - 30 = 180

3) Remove the 30*3 = 90 cases where numbers are odd: 180 - 90 = 90

4) Add the 15 cases included in both 2) and 3): 90 + 15 = 105