Originally Posted by

**TrueTears** First Question:

A 4 digit number ( with no repetitions) is to be formed from the set of numbers {0,1,2,3,4,5,6,7}. Find the probability that the number is even.

ok the way i tried it was this way:

_ _ _ _ <----There are 4 places u have to fill with numbers. last spot can occupy 4P1 ways. 1st spot can occupy 6P1 spots, (since 0 cant be on the 1st spot) and then the 2 spots left can be occupied in 6P2 ways.

However there can also be another set of possibility. the last spot can be filled 4P1 ways, but lets say 0 was picked for the last spot hence there are 7P1 ways of occupying the 1st spot and 6P2 for the last 2.

This gives: 4P1 x 6P1 x 6P2 + 4P1 x 7P1 x 6P2 = 1560 ways

then the total ways of occupying the 4 digits with no restrictions is 7P1 x 7P3 = 1470. 7P1 because for the 1st spot because it cant be 0. and the rest of the 3 spots have 7P3 ways of occupying.

however... there must be a mistake in my working since i got more ways of making a even number than the total number of ways with no restrictions lol.... can some1 just explain through how they would do it? Thanks