(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!


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  • $\begingroup$ Hi - how many of these 4-digit numbers end in 8? $\endgroup$ – Hans Engler Jun 9 '14 at 20:55

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.

  • $\begingroup$ I understood this, but how do I get the answer? Is there any formula I can use? $\endgroup$ – Kiara Jun 9 '14 at 17:51
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    $\begingroup$ You can multiply: $2 \times \text{number of 3-digit numbers from a pool of 5 numbers}$. $\endgroup$ – Namaste Jun 9 '14 at 17:54
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    $\begingroup$ This is the kind of problem that thinking it through is likely quicker, and possibly more likely to be correct, than trying to memorize and recall formulae! $\endgroup$ – Namaste Jun 9 '14 at 17:56
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    $\begingroup$ You're welcome, Kiara! $\endgroup$ – Namaste Jun 9 '14 at 18:00

$\bf{(II)}$ 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.


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