Si A et B sont disjoints : \( |A \cup B| = |A| + |B| \)
20 livres + 10 BD → 30 choix.
\( n \times m \)
3 pantalons × 4 chemises = 12 tenues.
\( n! = n(n-1)(n-2)...1 \)
5! = 120
\( n! \)
ABC → 6 arrangements possibles
\( A_n^p = \frac{n!}{(n-p)!} \)
20 élèves → podium de 3 : 20×19×18
\( \binom{n}{p} = \frac{n!}{p!(n-p)!} \)
Choisir 3 élèves parmi 10.
\( \binom{n}{p} = \binom{n-1}{p-1} + \binom{n-1}{p} \)
\( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} \)
\( \sum_{k=0}^{n} \binom{n}{k} = 2^n \)