# What is the sum (in base 10) of all the natural numbers less than 64 which have exactly three ones in their base 2 representation?

What is the sum (in base 10) of all the natural numbers less than 64 which have exactly three ones in their base 2 representation?

The numbers will have at most $6$ digits in their binary representation and let ﬁrst digits can be $0$ , hence all numbers can be made six digit numbers. The number of numbers with exactly $3$ ones $= 6C3= 6·5·4 6 = 20$. Each number has $3$ ones and $3$ zeros. So totally out of $20·6$ digits $20·3$ one’s are there. So there are exactly $20*3/6$ = $10$ one's in each place.

I don't know how I can proceed for here!

A natural number less than $64$ with exactly three ones in it's binary representation, can be written as the sum of three distinct powers of $2$, and vice-versa.
So what we are essentially doing, is adding up all numbers, that are a sum of three distinct powers of $2$. We know that there are six powers of $2$ below $64$. We'll put a notation : $\{a,b,c\}$ stands for $2^a + 2^b + 2^c$, where $a,b,c$ are distinct and between $0$ and $5$, both included. Now, we are adding up all such possibilities, whose number you have correctly calculated.
The point is, how many times does $a$ appear in these combinations, where $a$ is between $0$ and $5$? Suppose $a$ appears in a combination, then the other two can be chosen in $\binom 52 =10$ ways. Hence, the number of combinations in which $a$ appears is $10$.
Since this happens for all $a$, the answer should be $10(2^0 + ... + 2^5) = 630$.
There are ${6\choose3}=20$ numbers in $[0..63]$ having exactly three ones in their binary representation, makes $60$ ones in total. These ones are uniformly distributed over the six places in the binary representation, so that each of them creates ${1\over6}(1+2+\ldots+32)$ "worth" on average. It follows that the sum of all these numbers is $60\cdot{1\over6}\cdot63=630$.