Let's place the last digit: it must be 2 or 4 or 8 (3 possibilities)
It remainds 4 digits and the number of permutations fo 4 numbers is 4!=4*3*2*1=24
Thus there are 3*24=72 possibilities.
Answer A
If you do'nt believe run this programm
DIM n(5) AS INTEGER, i1 AS INTEGER, i2 AS INTEGER, i3 AS INTEGER, i4 AS INTEGER, i5 AS INTEGER, nb AS LONG, tot AS LONG tot = 0 n(1) = 1 n(2) = 2 n(3) = 4 n(4) = 7 n(5) = 8 FOR i1 = 1 TO 5 FOR i2 = 1 TO 5 IF i2 <> i1 THEN FOR i3 = 1 TO 5 IF i3 <> i2 AND i3 <> i1 THEN FOR i4 = 1 TO 5 IF i4 <> i3 AND i4 <> i2 AND i4 <> i1 THEN FOR i5 = 1 TO 5 IF i5 <> i4 AND i5 <> i3 AND i5 <> i2 AND i5 <> i1 THEN nb = ((((n(i1) * 10) + n(i2)) * 10 + n(i3)) * 10 + n(i4)) * 10 + n(i5) IF nb MOD 2 = 0 THEN tot = tot + 1 END IF END IF NEXT i5 END IF NEXT i4 END IF NEXT i3 END IF NEXT i2 NEXT i1 PRINT "tot="; tot END