In one section of a math test paper, a student answers 3 out of 4 questions in that section. 
In How many ways can this section be answered if different orders in which the questions are answered count as different ways.
My answer is 4*3*2 = 24     <<<---=== Correct !



In how many ways can 4 boys and 2 girls seat themselves in a row if
(a) the 2 girls are to seat next to each other
(b) the 2 girls are not to seat to each other
(c) the 2 girls are to be separated by 2 boys between them


for (a) I did 4!*2!*6 ( 4! is for the boys, 2! for the girls and 6 positions the girls can start at). This for me is a bit tricky for some reason.


        The correct solution is THIS:  we have 5 objects (4 boys and two glued girls, which we consider as one object)
                                       There are 5! = 120 permutations;  and the glued object can be in one of 
                                       the two states (Ann,Barb) or (Barb,Ann).
                                       So the total number of arrangements is  120*2 = 240.    ANSWER



for (b) I did 6! - 2!(6-2+1)! = 480

        Your answer number is correct.

        My standard logic for it is   6! - 240 = 720 - 240 = 480.    (the number 240 comes from part (a) ).



for (c) I did 4! * 2! * 3 (4! from the boys. 2! from the girls and 3 from the fact that there are only 3 cases of this that can be happened)

        Correct

        


in how many ways can 5-digit numbers be formed using the digits 0,1,3,4,5 if
(e) no repetition of digits is allowed
(d) repetition of digits is allowed



for (e) I did 4*4*3*2*1 = 96 (4 at first because zero is not allowed in the first digit of a number)

        Correct.



for (d) i did 5*5*5*5*5

        Incorrect.


        The correct counting is 4*5*5*5*5   (zero can not be leading digit !)