# Counting

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• May 18th 2010, 02:50 AM
jvignacio
Counting
Hey guys, im a little confused with these questions.. Any help would be much appreciated.

(a) How many 6 digit numbers are there?
(b) How many even 6 digit numbers are there?
(c) How many even 6 digit numbers with no repeated digits are there?
(d) How many even 6 digit numbers greater than 310,000 with no repeated digits are there?
• May 18th 2010, 03:23 AM
Debsta
Quote:

Originally Posted by jvignacio
Hey guys, im a little confused with these questions.. Any help would be much appreciated.

(a) How many 6 digit numbers are there?
(b) How many even 6 digit numbers are there?
(c) How many even 6 digit numbers with no repeated digits are there?
(d) How many even 6 digit numbers greater than 310; 000 with no repeated digits are there?

These all involve the multiplication principle. The best way to do them is to draw six boxes in a row (to represent the 6 digits), then fill each box with the number of possibilities and then multiply together.
Eg for Q1
9x10x10x10x10x10 = 900000
(The first box has 9 possibilities (it can't start with 0 otherwise it would be a five-digit number, the other boxes each have 10 possibilities)

Try the others for yourself and I can check them for you if you like.
• May 18th 2010, 04:06 AM
Debsta
Quote:

Originally Posted by Debsta
These all involve the multiplication principle. The best way to do them is to draw six boxes in a row (to represent the 6 digits), then fill each box with the number of possibilities and then multiply together.
Eg for Q1
9x10x10x10x10x10 = 900000
(The first box has 9 possibilities (it can't start with 0 otherwise it would be a five-digit number, the other boxes each have 10 possibilities)

Try the others for yourself and I can check them for you if you like.

How did you go?
• May 18th 2010, 04:07 AM
jvignacio
Quote:

Originally Posted by Debsta
How did you go?

Hey just reading your post.. I will try it now and post.. Thanks for the reply and help.
• May 18th 2010, 04:13 AM
jvignacio
Quote:

Originally Posted by Debsta
How did you go?

Ok so b) is...

Half of a) which is 900000/2 ... or I can do that multiplication principle but I can only get the first number which is 5x?x?x?x?x?
• May 18th 2010, 04:18 AM
Debsta
Quote:

Originally Posted by jvignacio
Ok so b) is...

Half of a) which is 900000/2 ... or I can do that multiplication principle but I can only get the first number which is 5x?x?x?x?x?

Even six digit numbers.
First digit can be 1 to 9 (ie 9 possibilities)
Second, third, fourth, fifth digits can each be 0 to 9 (ie 10 possibilities)
Last digit can be 0, 2,4,6,8 (ie 5 possibilities)

So:
9x10x10x10x10x5 = 450000 (which is half of the answer to (a) obviously)
• May 18th 2010, 04:22 AM
jvignacio
Quote:

Originally Posted by Debsta
Even six digit numbers.
First digit can be 1 to 9 (ie 9 possibilities)
Second, third, fourth, fifth digits can each be 0 to 9 (ie 10 possibilities)
Last digit can be 0, 2,4,6,8 (ie 5 possibilities)

So:
9x10x10x10x10x5 = 450000 (which is half of the answer to (a) obviously)

Oh right .. so i was correct about half of a).. but i was doing it the wrong way around.. Ok, let me try c) :)
• May 18th 2010, 04:28 AM
jvignacio
Quote:

Originally Posted by Debsta
How did you go?

hmm So for c) I got...

9x10x8x7x6x5 .. this ok?
• May 18th 2010, 04:30 AM
Debsta
Quote:

Originally Posted by jvignacio
hmm So for c) I got...

9x10x8x7x6x5 .. this ok?

Nearly!!
First digit (1 to 9) ie 9 possibilities
Second digit (you've already used a digit for the first spot so only 9 to choose from).
Then 8, then 7 etc.
So it should be:
9x9x8x7x6x5
• May 18th 2010, 04:34 AM
jvignacio
Quote:

Originally Posted by Debsta
Nearly!!
First digit (1 to 9) ie 9 possibilities
Second digit (you've alreadu used a digit for the first spot so only 9 to choose from.
Then 8, then 7 etc.
So it should be:
9x9x8x7x6x5

Oh yess I get it!! awesome.. let me try d) :P
• May 18th 2010, 04:44 AM
jvignacio
Quote:

Originally Posted by Debsta
How did you go?

ok for d) i got....

7x8x8x7x6x5 this correct?
• May 18th 2010, 04:58 AM
Debsta
Quote:

Originally Posted by jvignacio
ok for d) i got....

7x8x8x7x6x5 this correct?

mmmm don't think so.
This one is a bit trickier because if the first digit is 3 then the second digit has to be 1 or more BUT if the first digit is greater than 3 then the second digit CAN be 0.
I would suggest you use complements on this one.
That is, find the number of even numbers (no repeating digits) less than 310 000 and subtract from your answer to (c).
SO:
Consider the numbers which begin with 1:
Consider the numbers which begin with 2;
Consider the numbers that begin with 30....
Add these and subtract from (c)
• May 18th 2010, 05:15 AM
Debsta
Quote:

Originally Posted by Debsta
mmmm don't think so.
This one is a bit trickier because if the first digit is 3 then the second digit has to be 1 or more BUT if the first digit is greater than 3 then the second digit CAN be 0.
I would suggest you use complements on this one.
That is, find the number of even numbers (no repeating digits) less than 310 000 and subtract from your answer to (c).
SO:
Consider the numbers which begin with 1:
Consider the numbers which begin with 2;
Consider the numbers that begin with 30....
Add these and subtract from (c)

Actually I am now rethinking the solution for (c).
For the number to be even, it must end in 0, 2,4,6,8.
If it ends in 0, there are 9 ways to fill the first box.
BUT if it ends in 2,4,6,or 8, there are only 8 ways to fill the first box.
So we need to count how many end in 0, separately to those that end in 2,4,6,or 8.

End in 0:
?x?x?x?x?x1 (one possibility (ie zero) for the last box)
9x8x7x6x5x1 (then fill in the others from the beginning)

End in 2,4,6 or 8:
?x?x?x?x?x4 (4 possibilities at the end)
8x?x?x?x?x4 (8 possibilities for the first - can't be 0 or the end number)
8x8x?x?x?x4 (8 possibilities left for the second number)
8x8x7x6x5x4 (fill in the others)

So adding these (underlined) values will give the number of numbers you want (ie the 6-digit numbers that end in 0 and the numbers that end in 2,4,6,or 8, with no repeated digits.)
Sorry about the bum steer with (c) before.
• May 18th 2010, 05:15 AM
jvignacio
Quote:

Originally Posted by Debsta
mmmm don't think so.
This one is a bit trickier because if the first digit is 3 then the second digit has to be 1 or more BUT if the first digit is greater than 3 then the second digit CAN be 0.
I would suggest you use complements on this one.
That is, find the number of even numbers (no repeating digits) less than 310 000 and subtract from your answer to (c).
SO:
Consider the numbers which begin with 1:
Consider the numbers which begin with 2;
Consider the numbers that begin with 30....
Add these and subtract from (c)

wow thats heaps confusing to understand... Is there another way of doing this? I forgot to consider the numbers greater than 3 can have a zero on the second digit :(
• May 18th 2010, 05:22 AM
jvignacio
Quote:

Originally Posted by Debsta
Actually I am now rethinking the solution for (c).
For the number to be even, it must end in 0, 2,4,6,8.
If it ends in 0, there are 9 ways to fill the first box.
BUT if it ends in 2,4,6,or 8, there are only 8 ways to fill the first box.
So we need to count how many end in 0, separately to those that end in 2,4,6,or 8.

End in 0:
?x?x?x?x?x1 (one possibility (ie zero) for the last box)
9x8x7x6x5x1 (then fill in the others from the beginning)

End in 2,4,6 or 8:
?x?x?x?x?x4 (4 possibilities at the end)
8x?x?x?x?x4 (8 possibilities for the first - can't be 0 or the end number)
8x8x?x?x?x4 (8 possibilities left for the second number)
8x8x7x6x5x4 (fill in the others)

So adding these (underlined) values will give the number of numbers you want (ie the 6-digit numbers that end in 0 and the numbers that end in 2,4,6,or 8, with no repeated digits.)
Sorry about the bum steer with (c) before.

Crap!! thats so true! we didnt consider the numbers at the end were getting repeated at the beginning ... I understand this one now.. :)
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