What is P(0.6 ≤ z ≤ 2.12)?
2 answers:
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Answer: 151,200 combinations
Step-by-step explanation:
So we want to order the numbers 1, 2, .... , 9 in a:
such that the sum in each row is an odd number.
if we add 3 odd numbers we have an odd number.
if we add 2 even numbers, and an od number, we have an odd number.
where the rows are in this case (1, 2, 3), (4,5,6) and (7, 8, 9)
We have 4 even numbers and 5 odd numbers, so one combination can be:
Now, we for this option we can:
permute the numbers in each row: c =3*2*1 options per row and permute the rows, also c = 3*2*1
So only for this combination we have: P = (3*2)^4 possible ways of writing it.
Now, let's find the different combinations of numbers, so you can expect that we will find a lot of combinations, now let's construct them:
Suppose that in the first row we have two even numbers and 1 odd, then for the first even number we have 4 options, for the second 3 options, and for the odd number 5 options, so we have 4*3*5 combinations.
For the second row we also want two even numbers and one odd.
For the first even number we have 2 options, for the second one, and for the odd number we have 4 options, so we have 4*2*1 combinations.
In the last row we want the other 3 odd numbers, so we have 3*2*1 combinations, this adds to:
C = (4*3*5)*(3*2*1)*(4*2*1) = 2880 combinations.
If we add the fact that the rows can interchange positions, we have other 3*2*1 combinations:
C = 2880*3*2 = 16,800 and if now we add the fact that the numbers in the first two rows can permute (it is not the same (2, 4, 5) than (2, 5, 4) or (4, 2, 5)) (this happens because in these two rows we have different types of numbers, while in the third one the permutations are already included)
So the thing that can change in the first two rows is the position of the odd number, can be first, second or third, so we have other 3 permutations for the two first rows:
C = 16,800*3*3 = 151,200 which is the total number of different combinations.