We have deck of cards (52 cards). 10 random cards will be picked. How many chances are there, that atleast 1 of the random cards is "ace".
At first I tried to calculate how many different possibilities are there to pull 10 random cards without any extra conditions. According to formula nCk= n! / k!(n-k)!: I got 52! / 10! * 42!
Next I tried to calculate how many possibilities are to pull 10 cards from deck, that does not contain any "aces" I got 48! / 10! * 38!
Then I could subtract second answer from first and I would be left over with answer. Since the numbers are too huge to try on calculator, I wondered if my solutions works.
It's okay to give the answer as:
${52\choose 10} - {48\choose 10}$
That is an acceptable answer but if you want to caluculate it it is.
$\frac {52!}{42!10!} - \frac {48!}{38!10!} =$
$\frac {48!}{38!10!}(\frac {52*51*50*49}{42*41*40*39} - 1)=$
$\frac {39*....*48}{10!}(\frac {52*51*50*49}{42*41*40*39}-\frac {42*41*40*39}{42*41*40*39})=$
$\frac {43*44*....*48}{10!}(52*51*50*49 - 42*41*40*39)=$
.... factor terms out.
$\frac {(43*22*1*46*47*1)*6*8*9*10}{10!}(13*17*5*7 - 1*41*2*39)2*3*4*5*7=$
$(43*22*46*47)(13*17*5*7 - 41*2*39)$
And that can be plugged into a calculator.
But.... 1) NOBODY CARES!!!! The answer ${52\choose 10} - {48\choose 10}$ is good enough and
2) If you have a computer it comes with a calculator and the numbers aren't too high
$52! = 8.0658175170943878571660636856404e+67$
$\frac {52!}{42!} = 57407703889536000$
$\frac {52!}{42!10!} = 15820024220$
$48! = 1.2413915592536072670862289047373e+61$
$\frac {48!}{38!10!} = 6540715896$
and $\frac {52!}{42!10!} - \frac {48}{38!10!} = 9279308324$
.... which.... nobody cares. Unless you are doing probability ... in which case $\frac {{52\choose 10} - {48 \choose 10}}{52\choose 10}$ is easy to calculate with cancelation as $1 -\frac {48!42!}{52!38!}= 1- \frac {39*40*41*42}{49*50*51*52} = 1- \frac{41*6}{7*5*17}$