Binary less than 100

Firstly, you have to choose a 0 or 1 for each of the 7 spaces (as $2^{7}$ = 64 but, $2^{8}$ = 128 > 100). For a single 1, there are \ $boxed{7}$ choices (#1). For pairs: there are $7\choose2$ = $\boxed{21}$ choices (#2). Now for triples and quads I need to denote the spaces from A to G from left to right and I will need to use ${casework}$ . TRIPLES: If G = F = 1 (in binary), then there are 3 choices, namely (G,F, A/B/C). If G = 1, F $\neq$ 1, then there are $5\choose2$ = 10 choices. If G $\neq$ 1, F=1, then again there are $5\choose2$ = 10 choices. If G $\neq$ 1, F $\neq$ 1, then there are $5\choose4$ = 5 choices. Overall, for triples there are $\boxed{33}$ choices (#3). QUADS: If G=F=1, then A = 1 is the only option. If G = 1, F $\neq$ 1: there are $5\choose3$ = 10 choices; as with G $\neq$ 1, F $\neq$ 1. If G $\neq$ 1, F $\neq$ 1: there are $5\choose4$ = 5 choices. Overall, for quads there are $\boxed{26}$ choices (#4). In total, there are 7 + 21 + 33 + 26 = ${\color{#3D99F6}8}$ ${\color{#3D99F6}7}$ choices

We will prove our answer by looking at the complement.

Looking at binary, we have the largest value less than 100 to be $1100100$

There will be $\dbinom{7}{5}+\dbinom{7}{6}+\dbinom{7}{7}=29$ numbers that have a digit sum greater than 4.

Now we have to look at numbers that exceed 100 but have a digit sum less than 4

There will be 4 numbers that are less then 100 and have a digit sum less than 4 (1100100, 1100011, 1100001, 1100010), so we'll put these aside for now.

Looking at the numbers, we have $\dbinom{5}{1}+\dbinom{5}{2}=15$

However, we already decided that 4 will work, so we subtract 4, $15-4=11$

Totaling our numbers that don't work we have 11+29=40

Since we have 127 numbers between 127 and 1, and 40 that don't work, we have

$127-40=87$ numbers that work.

Note that $100 = 1100100_{2}\\$ Counting backwards from 100, note that the first number with a digit sum of 5 or more is $101 1111_{2}\\$ From the last sequence of $1 1111$ we exclude those numbers with a digit count of 5 or 4. ${5 \choose 5} + {5 \choose 4} = 1 + 5 = 6\\$ The next lowest number with a digit count of 5 or more is $011 1111_{2}\\$ Again, we exclude the $6$ numbers with a digit count of 5 or 4 from the low order sequence. $\\$ There is one final sequence with a digit count of 5 - $0011111_{2}\\$ Hence the answer is $100-(6+6+1)=\boxed{87}$