Let us solve the given function
$F = AB’C + A’BC + ABC’ + A’B’C + AB’C'$
$F = AB'(C + C') + A'C(B + B') + ABC'$
$F = AB'.1 + A'C.1 + ABC' [C + C' = 1, B + B '= 1 ]$
$F = AB' + A'C + ABC'$
$F = A(B' + BC')+ A'C$
$F=A( ( B'+B ).( B' + C' ) ) + A'C \:\:\:\:\:[\because \text{Using Distributive law}: A + B.C = (A + B).(A + C)]$
$F = A.1.( B' + C' ) + A'C \:\:\:\:\:\:\:\:[\because B' + B = B + B' = 1 ]$
$F = A.(B' + C' ) + A'C$
$F = AB' + AC' + A'C$
So, the correct answer is $(B)$.