How many numbers less than $100$ can be expressed as a sum of distinct factorials? How many numbers less than $100$ can be expressed as a sum of distinct factorials?
Example:
a) $4 = 2! + 2!$
b) $3 = 2! + 1!$
 A: Lemma(I): For every positive integer $n$ we have: 
$$ 
1! + 2! + ... + (n-1)! < n! 
\ \ 
$$

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 $i=1, 2, 3, 4$.  

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.




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

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

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 $i=0, 1, 2, 3, 4$.  

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. 
