Question 48934: how many numbers of five different digits, each number to contain 3 odd and 2 even digits, can be formed from the digits 1,2,3,4,5,6,7,8 and 9?Thanks!
Found 2 solutions by longjonsilver, 303795:
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! you need to build up the solution.
Even numbers: 2,4,6,8
Odd numbers: 1,3,5,7,9
The 5 digit number: XXXYY has to have 3 odds (the X's). So, lets look at just those:
Pick the first odd number. How many possibles are there? Answer 5 (1,3,5,7,9)
Now pick the second odd number. How many possibles are there now? Answer 4 (since one of them was picked to be the first number.
Similarly, the third odd number... there are 3 possibles.
So altogether, there are 5*4*3 --> 60 3-digit variations from 5 odd numbers.
Now for the evens. For a 2-digit number, the first of the 2 even numbers could be 1 of 4 numbers: answer = 4
The final even number must be one of the remaining 3 evens: answer = 3
So, altogether there are 4*3 --> 12 2-digit variations from 4 even numbers.
And now we combine them... we know how many odd permutations there are (60) and even permutations (12). Combined, there are 60*12 --> 720 permutations of 3 odd and 2 even.
As you can see... at no point did i use any "formula". I just worked it out from "first principles" and "common sense"
Hope this helps.
jon.
Answer by 303795(602)  (Show Source):
You can put this solution on YOUR website! Look at the solution by Longjonsilver.
His solution shows that there are 720 combinations but he has assumed that the three odd numbers go first then the two even numbers.
There are 10 different ways in which the 3 odd and 2 even numbers can be distributed.
O O O E E
O O E E O
O O E O E
O E O E O
O E O O E
O E E O O
E O O E O
E O O O E
E O E O O
E E O O O
This brings the total number of solutions to 720 * 10 = 7200 which meet the original conditions.
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