# Permutation and Combination

• Feb 8th 2009, 04:30 AM
azuki
Permutation and Combination
Find how many 5-digit numbers can be formed from digits 1,2,3,4,5, if
a) the numbers form must be odd,
b) the numbers formed must be divisible by 4,
c) the odd digits must occupy even positions (i.e. 2nd and 4th) and the even digits must occupy odd positions (i.e. 1st, 3rd, and 5th).
• Feb 8th 2009, 06:12 AM
galactus
For the first one, in order to be odd the last digit must be 1,3 or 5.

Place the 1 at the end and arrange the other 4 digits in 4!=24 ways.

Place the 3 at the end and arrange the other 4 digits in 4!=24 ways.

Same for the 5

24+24+24=72 different numbers.
• Feb 15th 2009, 12:23 PM
Jhevon
Quote:

Originally Posted by azuki
b) the numbers formed must be divisible by 4,

a number is divisible by 4 if its last 2 digits are divisible by 4.

so of all the even numbers, how many end with 2 digits that are divisible by 4?

Quote:

c) the odd digits must occupy even positions (i.e. 2nd and 4th) and the even digits must occupy odd positions (i.e. 1st, 3rd, and 5th).
this seems like a strange question. you will always have at least one odd digit occupying an odd position.