Permutations

carebear

New member
Joined
Aug 30, 2010
Messages
45
How many different numbers can be formed from the five digits 1, 2, 3, 4, 5
a) if the odd digits must occupy the odd places?

I get: 3 x 2 x 2 x 1 x 1 = 12 ways....which I believe is correct but I do not understand the next question......

b) if the odd digits and the even digits must both be in ascending order? I see soooo many different cases so do not know where to start.

My teacher told me that the answer is 10 but I do not understand how she got it.

Please help!!
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,489
10 is correct:
24 appears 4 times
2-4 appears 3 times
2--4 appears 2 times
2---4 appears 1 time

Get my drift?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, carebear!

How many different numbers can be formed from the five digits 1, 2, 3, 4, 5

a) if the odd digits must occupy the odd places?

I get: 3 x 2 x 2 x 1 x 1 = 12 ways . . . which I believe is correct but I do not understand the next question . . .


b) If the odd digits and the even digits must both be in ascending order?

My teacher told me that the answer is 10.

We have a five-digit number: .\(\displaystyle \square\;\square\;\square\;\square\;\square\)

The even digits must be in the order 2,4.
The odd digits must be in the order 1,3,5.

Choose two spaces for the two even digits.
. . \(\displaystyle \text{There are: }\:_5C_2 \:=\:{5\choose2} \:=\:\frac{5!}{2!\,3!} \:=\:10\,\text{ choices.}\)

Once you choose the two spaces, the two even digits are inserted in 2-4 order.
Then the three odd digits are inserted in the remaining spaces in 1-3-5 order.

So there are 10 numbers with odds and evens in ascending order.


In case you're skeptical, here they are:

. . \(\displaystyle \begin{array}{ccccc}2&4&1&3&5 \\ 2&1&4&3&5 \\ 2&1&3&4&5 \\ 2&1&3&5&4 \\ 1&2&4&3&5 \\ 1&2&3&4&5 \\ 1&2&3&5&4 \\ 1&3&2&4&5 \\ 1&3&2&5&4 \\ 1&3&5&2&4 \end{array}\)
 
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