Using the digits 2, 3, 4, 7 and 8, Carlos will form five-digit positive integers. Only the digit 2 can be used more than once in any of Carlos’ five-digit integers. How many distinct fivedigit positive integers are possible?
+128474 Using 2 once = all arrangements of 5 different digits = 5! = 120
Using 2 twice = choose any two positions for the 2s, = C(5,2).....choose any 3 of the remaining 4 digits =C(4,3)...and arrange them in 3! ways = C(5,2) * C(4,3) * 3! = 240
Using 2 three times = choose any 3 poisitions for the 2s = C(5, 3) ....choose any 2 of the 4 remaining digits = C(4,2) and arrange these in 2! ways = C(5,3) * C(4,2) * 2! = 120
Using 2 four times = choose any 4 of 5 positions for the 2s = C(5,4) and for the other position, choose any 1 of the remaining 4 digits...C(5,4) * C(4,1) = 20
Using all 5 2s = 1
120 + 240 + 120 + 20 + 1 =
501 numbers
