start from constraint given that w stands in front of y
method one - listing options
_y _ _ , 1*1*2*1=2
_ _ y _ , 2*1*1*1 OR 2*1*1*1 ==4
_ _ _ y , 3*2*1*1=6
total ways' count is (2+4+6)=12

method two - counting
fix y in the second place and count ways, 1 for w and 2C1 for other alphabets 1*2C1=2
fix y in the third place ~, 2 for w and 2C1 for other alphabets 2*2C1=4
fix y in fourth place ~, 3! ways for three alphabets=6
total ways's count is (2+4+6)=12

geet_ge wrote:In how many ways can you arrange 4 alphabets(w,x,y & z) in a straight line so that w comes before y?

Simple version...

Start as permutation: 4 * 3 * 2 * 1 = 24 ways to write the letters. Then, w will be in front of y exactly half of those times.

24/2 = 12 ways

Hi

If W in the first position then total no ways=3!=6 ways
if W in the second position total ways=3p2 -1=5 ways
if w is in 3rd position only =1 way

total=6+5+1=12 ways.