Using the digits 2, 3, 4, 7 and 8, Carlos will form five-digit positive integers. Only the digits 2 and 3 can be used more than once in any of Carlos’ five-digit integers. How many distinct fivedigit positive integers are possible?
In cases such as this one, it is often a lot easier to compute the combinations first, then convert them into permutations as follows:
(22222, 22223, 22224, 22227, 22228, 22233, 22234, 22237, 22238, 22247, 22248, 22278, 22333, 22334, 22337, 22338, 22347, 22348, 22378, 22478, 23333, 23334, 23337, 23338, 23347, 23348, 23378, 23478, 33333, 33334, 33337, 33338, 33347, 33348, 33378, 33478) >>Total = 36 combinations.
These 36 combinations break down as follows:
2 Ccombinations x 1 Permutation each == 2P
8 C x 5P ==40P
2 C x 10P ==20P
12 C x 20P ==240P
3 C x 30P ==90P
8 C x 60P ==480P
1 C x 120P==120P
Total combinations == 36
Total permutations == 992
+37153 Start with not using 478
then you have 2 x 2 x 2 x 2 x 2 = 32 combos
then you can have ONE 8 or ONE 7 or ONE 4 each in one of the 5 positions = 15 the other 4 positions 2 x 2 x 2 x 2 possible
15 x 16 = 240
then you can have TWO of the 4 7 8 47 74 78 87 48 84 5 C 2 x 6 = 60
the other three positions would be 2 x 2 x 2 = 8
60 x 8 = 480
you can have all of 4 7 8 in 6 orders x 5 C3 = 60 the other two positions filled by 2 x 2
4 x 60 = 240
Summing the reds = 992 possible numbers ( same answer guest found !)
