Stephanie M. answered • 07/07/15
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Imagine first selecting four digits to be part of the four-digit number, then arranging those digits.
(a)
We will first select four digits to be part of our number. Since the even digits must be side by side, we'll select 2 and 4. How many ways are there to select the remaining two digits from the four odd numbers?
(4 choose 2) = 4!/(2!(4-2)!) = (4×3×2×1)/(2×1(2×1)) = (4×3×2×1)/4 = 3×2×1 = 6 ways
Those ways, by the way, are (3, 5), (3, 7), (3, 9), (5, 7), (5, 9), and (7, 9).
Once we have those four digits, we'll need to arrange them. Since the 2 and 4 must be next to each other, we can view them as a single unit. So, how many ways are there to put three things in an order?
3×2×1 = 6 ways
For example, assume we have 2, 3, 4, and 5. The ways to arrange (24), 3, and 5 are 2435, 2453, 3245, 5243, 3524, and 5324.
Finally, wherever the 2 and 4 are, they could be in the order 24 or in the order 42. So there are 2 ways to arrange the 2 and the 4.
Putting all that together, there are a total of 6×6×2 = 72 possible four-digit numbers.
(b)
Again, our number must include 2 and 4, since alternating odd and even digits means two odd digits and two even digits. And again, there are 6 ways to choose the remaining two odd digits.
One of the two even numbers must occur first in our four-digit number. The 2 and 4 will either occur in the order 2_4 or 4_2. So, there are 2 ways to arrange the even numbers.
Similarly, one of the two odd numbers must occur first. If we chose 3 and 5, for example, their order must be 3_5 or 5_3. So, there are 2 ways to arrange the odd numbers.
Finally, our number could either start with an even number (EOEO) or start with an odd number (OEOE). So, there are 2 ways to arrange the groups of odd and even numbers.
Putting that all together, we get a total of 6×2×2×2 = 6×8 = 48 possible four-digit numbers
