Find how many of these $4$-digit numbers are even.

Question

(a)

(i) Find how many different $4$-digit numbers can be formed from the digits $1, 3, 5, 6, 8$ and $9$ if each digit may be used only once. I did this; the answer is $360$; I used $\frac{n!}{(n-r)!}$

(ii) Find how many of these $4$-digit numbers are even. Now I can't do this....please help...Fast! Tomorrow's my exam!

Answer 1

Only $6, 8$ are even digits among your choices. To be even, the unit's digit must be even. That gives you two choices for the unit's digit, and you can take it from there.

That means we have $$2\cdot 5\cdot 4\cdot 3 = 120$$ distinct ways of creating, using the given pool of letters, 4-digit number that are even.

Answer 2

$\bf{(II)}$ Here I am assuming that repetition is not allowed.

If $4$ digit formed no. is even , Then last digit (means unit digit) must be even.

this can be done in $2$ ways because there are only $2$ even digit.

Now $10^{th}\;,100^{th}\;,1000^{th}$ placed can be filled in $5\;,4\;,$ and $3$ ways respectively.

So Total no. of ways of forming $4$ Digit no. is $ = 2\times 5 \times 4 \times 3 = 120$ ways.