With the five numbers 1-5 how many such 5-digit numbers are : Quant Question Archive [LOCKED]
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# With the five numbers 1-5 how many such 5-digit numbers are

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With the five numbers 1-5 how many such 5-digit numbers are [#permalink]

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15 Sep 2007, 15:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

With the five numbers 1-5 how many such 5-digit numbers are possible in which even numbers are not adjacent

1.120
2.100
3.96
4.72
5.48
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15 Sep 2007, 16:33
Total arrangements for five items = 5!

Total arrangements for five items when two items are adjacent = 4!*2

Total arrangements for five items when two items are not adjacent = 5! - 4!*2 = 120 - 48 = 72

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15 Sep 2007, 18:17
KillerSquirrel wrote:
Total arrangements for five items = 5!

Total arrangements for five items when two items are adjacent = 4!*2

Total arrangements for five items when two items are not adjacent = 5! - 4!*2 = 120 - 48 = 72

Agree....

- Brajesh
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15 Sep 2007, 19:32
I dont understand why the adjacent numbers is 4!*2?

I mean i get the 4! part..dont know how you got the 2?

KillerSquirrel wrote:
Total arrangements for five items = 5!

Total arrangements for five items when two items are adjacent = 4!*2

Total arrangements for five items when two items are not adjacent = 5! - 4!*2 = 120 - 48 = 72

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15 Sep 2007, 19:54
fresinha12 wrote:
I dont understand why the adjacent numbers is 4!*2?

I mean i get the 4! part..dont know how you got the 2?

KillerSquirrel wrote:
Total arrangements for five items = 5!

Total arrangements for five items when two items are adjacent = 4!*2

Total arrangements for five items when two items are not adjacent = 5! - 4!*2 = 120 - 48 = 72

Two digits can be arranged in 2 ways.

- Brajesh
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15 Sep 2007, 20:46
studentnow wrote:
With the five numbers 1-5 how many such 5-digit numbers are possible in which even numbers are not adjacent

1.120
2.100
3.96
4.72
5.48

question is not clear whether the repetation is allowed or not.

if no repetation, then it is = 5! - 4! x 2 = 72
if repetation is allowed, it would be very different.
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15 Sep 2007, 21:42
studentnow wrote:
With the five numbers 1-5 how many such 5-digit numbers are possible in which even numbers are not adjacent

1.120
2.100
3.96
4.72
5.48

The more I do em, the easier combinatorics get

Ans D.

1,2,3,4,5. Adjacent meaning "next to eachother."

Lets find the total number of ways we can position these 5 numbers. 5!=120.

Now lets find the number of ways that 4 and 2 ARE next to each other.

Make 4 and 2 one unit. 4-2,1,2,5 --> 4!. We aren't done yet. We must multiply 4! by 2 because we have yet to consider 2-4,1,3,5

4!*2=48. 120-48=72.
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30 Sep 2007, 19:18
OlgaN wrote:
Easier way...
2 and 4 can be arrange so that they wiil not be adjacent only in 12 ways
other three numbers can be arranged in 3! ways
Thus 3!*12=72

That is fast. But how did u get 12? did u have to skectch..
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30 Sep 2007, 20:37
KillerSquirrel wrote:
Total arrangements for five items = 5!

Total arrangements for five items when two items are adjacent = 4!*2

Total arrangements for five items when two items are not adjacent = 5! - 4!*2 = 120 - 48 = 72

nice approach.. agreed with D!
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30 Sep 2007, 21:11
killer squirrel did this in a cool way. i didn't think of it first instinct.

i did this.
o=odd
e=even

the numbers can be arranged in these ways in which the even numbers are not side by side.

oeoeo
ooeoe
eoooe
eoeoo
oeooe
eooeo

and this results in 12 possible numbers for each string of o's and e's

3! * 2! = 12

and there are 6 different strings so 6 * 12 = 72.
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30 Sep 2007, 22:06
# of 5 digit numbers = 5! = 120

Possibilities where evens are next to each other:
2,4,_,_,_ -> 2 * 3! = 12 ways
_,2,4,_,_ -> 2 * 3! = 12 ways
_,_,2,4,_ -> 2 * 3! = 12 ways
_,_,-,2,4 -> 2 * 3! = 12 ways

Total 48 ways.

So # of 5-digit numbers where even digits are not in adjacent places = 120-48 = 72
30 Sep 2007, 22:06
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