Hello,
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
