How many even 5-digit numbers can be formed from the numbers 1, 2, 4, 7, 8 if no digit is repeated in a number? a. 72 b. 120 c. 124 d. 128

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

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .