Birhdays: find the probabilities for the various configurations of the birthdays of 22 people Let S,D,T,Q stand for simple,double,triple and quadruple, respectively:
So, for example: the probabilities of 22 simple birthdays(22 person have birthdays in different days) are $ P(22S) = (365!/343!)/365^-22 $
$P(20S + 1D) = (365!/344!1!)*(22!/20!2!)/365^-22 $ : in this formula we have use twice multinomial coeficient for dividing into groups:birthday groups(one group simple, second group double) and people's group(20 persons have simple birthdays, 2 persons double)
The same principle was used for P(18S + 2D) and P(16S + 3D)
BUT:
$P(14S + 4D) = (365!/347!)*(22!/14!2!2!2!2!)/365^-22$ : and now i am completely lost. Here has been used another principle for birthdays( the part: 365!/347!)
But why the author has used different principles?And can someone please explain the second principle?
And also the second principle was used, for exaple:
$P(19S + 1T) = (365!/345!)*(22!/19!3!)/365^-22$
$P(15S + 2D + 1T) = (365!/347!)*(22!/15!2!2!3!)/365^-22$
Thank you
 A: I'm getting a different answer for $P(14S+4D)$ and $P(15S+2D+1T)$. I'll explain my reasoning for the first one. The second is similar.
$P(14S+4D)$: The idea is we have to count the ways to choose the days for $14S$ and $4D$, then, for each such choice, count the ways we can arrange the $22$ people to have those chosen days.
One way to do this is a step-by-step approach:
\begin{eqnarray*}
&& \text{$\binom{365}{14}$ ways to choose the $14$ single days} \\
&& \text{$\binom{365-14}{4} = \binom{351}{4}$ ways to choose the $4$ double days} \\
&& \text{$\binom{22}{14}$ ways to choose the $14$ people for single days} \\
&& \text{$14!$ ways to arrange these $14$ people} \\
&& \text{$\binom{22-14}{2} = \binom{8}{2}$ ways to choose the $2$ people for first double} \\
&& \text{$\binom{8 - 2}{2} = \binom{6}{2}$ ways to choose the $2$ people for second double} \\
&& \text{$\binom{6-2}{2} = \binom{4}{2}$ ways to choose the $2$ people for third double} \\
&& \text{$\binom{4-2}{2} = \binom{2}{2}$ ways to choose the $2$ people for fourth double} \\
&& \\
&& \therefore\quad\text{#ways to get $14S$ and $4D$} = \binom{365}{14} \binom{351}{4} \binom{22}{14} 14! \binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{2}{2} \\
&& = \dfrac{365!\;22!}{14!\;347!\;4!\;2!\;2!\;2!\;2!}.
\end{eqnarray*}
$$\\$$
Another way to see it uses a multinomial coefficient:
Of the $365$ days we have $14$ S's, $4$ D's and $347$ 0's. So the number of day selections is the multinomial coefficient:
$$\dfrac{365!}{14!\;4!\;347!}.$$
Next, we have $22$ people to arrange among each such day selection but there are $4$ lots of pairs of people having the same day number. So the number of people arrangements is:
$$\dfrac{22!}{2!\;2!\;2!\;2!}.$$
The total number of $14S+4D$ arrangements is the product of these two amounts, which is the same answer as above.
