a) Find the number of 8-digit numbers that can be formed.
The distinguishable arrangements of 2, 2, 4, 5, 7, 7, 7, 7
If they were all distinguishable there would be 8!.
But the 4 7's are indistinguishable, so we divide by 4!.
Also the 2 2's are indistinguishable, so we divide by 2!.
So that's 8!/(4!2!) = 840 ways
b) Find the number of 7-digit numbers that can be formed when all 7’s are included.
The digits are either
Case 1: distinguishable arrangements of 2,2,4,7,7,7,7
[That's 7!/(4!2!) = 105]
or Case 2: distinguishable arrangements of 2,2,5,7,7,7,7
[That's also 7!/(4!2!) = 105]
or Case 3: distinguishable arrangements of 2,4,5,7,7,7,7
[That's 7!/4! = 210]
Total for all 3 cases: 105+105+210 = 420
c) Find the number of 7-digit numbers that can be formed.
In addition to the 420 in part (b), the others are when the numbers are
the distinguishable arrangements of 2,2,4,5,7,7,7.
That's 7!/(2!3!) = 420
Total = 420+420 = 840
Edwin