How many numbers less than $100$ can be expressed as a sum of distinct factorials?

Question

How many numbers less than $100$ can be expressed as a sum of distinct factorials?

Example:

a) $4 = 2! + 2!$

b) $3 = 2! + 1!$

Answer 1

Lemma(I): For every positive integer $n$ we have: $$ 1! + 2! + ... + (n-1)! < n! \ \ $$

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There are $ \color{Green}{15} = \color{Green}{16} \color{Red}{-1} = \color{Green}{2^4} \color{Red}{-1}$.

$$n= \varepsilon_1 (1!) + \varepsilon_2 (2!) + \varepsilon_3 (3!) + \varepsilon_4 (4!) ; $$

where $\varepsilon_i \in \{0,1\}$ for $i=1, 2, 3, 4$.

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Because for every $\varepsilon_i$ we have two choices.
The $\color{Red}{-1}$ appears because the case $\varepsilon_1=\varepsilon_2=\varepsilon_3=\varepsilon_4=0$ is not allowed;
as @Professor Vector has been mentioned.

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If $0!$ permited to join the sum then:

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Lemma(II): For every integer $3 \leq n$ we have: $$ 0! + 1! + 2! + ... + (n-1)! < n! \ \ $$

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There are $ \color{Green}{23} = \color{Green}{24} \color{Red}{-1} = \color{Green}{3.2^4} \color{Red}{-1}$.

$$n= \varepsilon_0 (0!) + \varepsilon_1 (1!) + \varepsilon_2 (2!) + \varepsilon_3 (3!) + \varepsilon_4 (4!) ; $$

where $\varepsilon_i \in \{0,1\}$ for $i=0, 1, 2, 3, 4$.

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Because for each of $\varepsilon_2, \varepsilon_3, \varepsilon_4$ we have two choices;
and we have three coices for choosing $\varepsilon_0, \varepsilon_1$;
i.e. $(\varepsilon_0, \varepsilon_1)= (0,0) \ \ \ \text{or} \ \ \ (0,1) \ \ \ \text{or} \ \ \ (1,1) . $
[Notice that the two pairs $(\varepsilon_0, \varepsilon_1)= (0,1)$ and $(\varepsilon_0, \varepsilon_1)= (1,0)$ are the same. ]
The $\color{Red}{-1}$ appears because the case $\varepsilon_0=\varepsilon_1=\varepsilon_2=\varepsilon_3=\varepsilon_4=0$ is not allowed;
as @Professor Vector has been mentioned.