Combinatorics question, if $n \geq 3$ then $B(n) < n!$ proof

Question

The question says Prove that if $n \geq 3$ then $B(n) < n!$

I am reading the solution in the book and I having a hard time understanding it

The question says Prove that if $n \geq 3$ then $B(n) < n!$.

The book proved this by induction. I understand the base case, $B(3)=5$ and $3!=6$ So $5<6$ is true for the base case.

Now to prove that this is true for $n+1$,

$$B(n+1) = \sum_{i=0}^n {n \choose i}B(i) < \sum_{i=0}^n {n \choose i} i!$$

I don't understand how they got $\sum_{i=0}^n {n \choose i} i!$ and I can't follow anything from there.

Answer 1

Since $B(i)\leq i!$ for $i=0,1,\dots n$, by the (strong) induction hypothesis, and $B(i)<i!$ when $i\geq 3$, you have that if $n\geq 3$ then:

$$\sum_{i=0}^n \binom{n}{i}B(i)\leq \sum_{i=0}^{n} \binom{n}{i}i!$$

For the rest of the proof: $\binom{n}{i}i!=\frac{n!}{(n-i)!}\leq n!$

So $$\sum_{i=0}^{n} \binom{n}{i}i!\leq (n+1)n!=(n+1)!$$