TheDonkey
Eat my bearrrrrrrrrrr, Tonighttt
+163|5708|Vancouver, BC, Canada
So basically, I deperately need to get 73% in this online math course I'm taking to get into university. I'm getting 90's on thier assignments (which are easy) but I need help actually understanding/applying the knowledge to different types of questions.

They posted their older exams so I'm working on solving those so I can get as many different styles of question figured out as I can. But I just can;t figure this one out,

Consider the numner 11111. How many different 7-digit numbers can be formed by inserting 2 and 3 in between the 1's if 2 and 3 can not be adjacent to each other? For example, an allowed number is 1213111. A number like 1231111 is not allowed. The 7-digit number must begin and end with 1.

The way I looked at is was there was one way to get the first number (just 1), 3 possible ways to get the second (1, 2, or 3), and 2 possible ways to get the third (1, 2/3 (whichever hasn't been used), then just one way to get the rest. But then I get an answer of 6 and that doesn't take into account that 2 and 3 can't touch....

Hallppp D:

Correct answer is 12.
Spamtheban
Undsiputed BF2s FIFA champion of all time
+132|4833|Stoke
12
13urnzz
Banned
+5,830|6488

48÷2(9+3) = 12
presidentsheep
Back to the Fuhrer
+208|5952|Places 'n such
nPr?
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
TheDonkey
Eat my bearrrrrrrrrrr, Tonighttt
+163|5708|Vancouver, BC, Canada

Spamtheban wrote:

12
...

burnzz wrote:

48÷2(9+3) = 12
no
TheDonkey
Eat my bearrrrrrrrrrr, Tonighttt
+163|5708|Vancouver, BC, Canada

presidentsheep wrote:

nPr?
With which numbers though?
4· 3 = 12.
presidentsheep
Back to the Fuhrer
+208|5952|Places 'n such
No idea, its n objects r at a time, I'm sure there's a way to do it but I cant think at the minute.
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
Jay
Bork! Bork! Bork!
+2,006|5349|London, England
2121212
3131313
1212121
1313131

3121212
2131212
2121312
2121213

3131212
2131312
2121313
3121213

3131312
2131313
3121313
3131213

1312121
1213121
1212131

1313121
1213131
1312131

That's just for one set. You're looking at quite a number of them but I can't be assed to think up a formula for you, sorry. It's 4 sets of one type, and 3 sets of the other. You'll have to pick out the overlapping ones too.
And whoever wrote that question needs his teeth knocked out because it's less than pointless.


I think that's all of them: 22

Reading is fundamental (starting with 1's, duh): 8

Last edited by Jay (2011-06-01 12:58:29)

"Ah, you miserable creatures! You who think that you are so great! You who judge humanity to be so small! You who wish to reform everything! Why don't you reform yourselves? That task would be sufficient enough."
-Frederick Bastiat
Ilocano
buuuurrrrrrppppp.......
+341|6658

1010101 + 1101011 = 2^3 + 2^2 = 12

The zero position is the only place that you can place 2's and 3's.
Jay
Bork! Bork! Bork!
+2,006|5349|London, England

Ilocano wrote:

1010101 + 1101011 = 2^3 + 2^2 = 12

The zero position is the only place that you can place 2's and 3's.
Which set did I miss?

Oh, I missed that the digit can be 1 too Oh well.

Last edited by Jay (2011-06-01 13:00:48)

"Ah, you miserable creatures! You who think that you are so great! You who judge humanity to be so small! You who wish to reform everything! Why don't you reform yourselves? That task would be sufficient enough."
-Frederick Bastiat
Ilocano
buuuurrrrrrppppp.......
+341|6658

Jay wrote:

Ilocano wrote:

1010101 + 1101011 = 2^3 + 2^2 = 12

The zero position is the only place that you can place 2's and 3's.
Which set did I miss?

Oh, I missed that the digit can be 1 too Oh well.
First at last have to be 1's.
Jay
Bork! Bork! Bork!
+2,006|5349|London, England

Ilocano wrote:

Jay wrote:

Ilocano wrote:

1010101 + 1101011 = 2^3 + 2^2 = 12

The zero position is the only place that you can place 2's and 3's.
Which set did I miss?

Oh, I missed that the digit can be 1 too Oh well.
First at last have to be 1's.
I would hope it would have a limit like that, because I just worked out about 30 variations using multiple 2's and 3's before I got bored
"Ah, you miserable creatures! You who think that you are so great! You who judge humanity to be so small! You who wish to reform everything! Why don't you reform yourselves? That task would be sufficient enough."
-Frederick Bastiat
Ilocano
buuuurrrrrrppppp.......
+341|6658

I loved permutations.  Get banned in Vegas.
presidentsheep
Back to the Fuhrer
+208|5952|Places 'n such
So if the first and last have to be 1s then you only really have 5 numbers to play with 111 2 and 3.
5P2 is 20?
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
Jay
Bork! Bork! Bork!
+2,006|5349|London, England
I would rather do differentials all day than deal with tedious shit like that
"Ah, you miserable creatures! You who think that you are so great! You who judge humanity to be so small! You who wish to reform everything! Why don't you reform yourselves? That task would be sufficient enough."
-Frederick Bastiat
DefCon-17
Maple Syrup Faggot
+362|6147|Vancouver | Canada
I thought you needed help with menstruation.
Ilocano
buuuurrrrrrppppp.......
+341|6658

Jay wrote:

I would rather do differentials all day than deal with tedious shit like that
Permutations for gambling.  Differentials for engineers and scientists.
TheDonkey
Eat my bearrrrrrrrrrr, Tonighttt
+163|5708|Vancouver, BC, Canada

Ilocano wrote:

Jay wrote:

I would rather do differentials all day than deal with tedious shit like that
Permutations for gambling.  Differentials for engineers and scientists.
Going into engineering. Still need it for high-school level math.

Thanks guys! Karmas to all.

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