Too Many 2s

The expanded summation of powers of two of a number is another way of describing how it is written in base 2. Let's use the given example, ${ 101 }_{ 10 }={ 1100101 }_{ 2 }$ Notice that the coefficients given in the powers of two summation form of 101 are mirrored in it's base 2 representation - the 0th, 2nd, 5th, and 6th digits are all 1, starting with the least significant digit as the 0th power and moving left.

Writing powers of two in base 2 is rather simple. In the same way that ${ 10 }^{ x }$ is a 1 followed by x zeroes in base 10, ${ 2 }^{ x }$ is a 1 followed by x zeroes in base 2. ${ 2 }^{ 222 }$ must then be a one followed by 222 digits, and is 223 digits long.

Once again in the same way that ${ 10 }^{ x }$ is the lowest x+1 digit number in base 10, $2^{ x }$ is the lowest x+1 digit number in base 2. Since we're considering only numbers less than ${ 2 }^{ 222 }$ , we are therefore considering every number with 222 digits or less in base 2.

Let's imagine a series of 222 zeroes written in a row. If we can "flip" some of them to ones, we can represent every number from $0$ to ${ 2 }^{ 222 }-1$ in binary. We know from earlier that if we flip any 4 unique zeroes to ones, then the number will have four terms in its summation form.

The number of different ways we can choose 4 zeroes to flip from 222 is given by the binomial coefficient $nCr(222, 4)=\frac { 222! }{ 4!(222-4)! } =\frac { (219)(220)(221)(222) }{ (1)(2)(3)(4) } =\boxed{98491965}$

To generalize, the number of integers less than ${ 2 }^{ a }$ that have $b$ terms when expressed as a sum of distinct powers of two is simply $nCr(a, b)$ .

Note: This perhaps isn't the most direct path of finding the solution, but I like it because it's relatively easy to understand when solved with the binary digits analogy.

treat the numbers as binary forms then numbers under 2^222 will have 222 digits in binary form of which as per question there must be 4 distinct powers of 2 so now for powers of 2 we take 1,1,1,1 and 218 0s
then permuting them weget 222! /(218!4!) which gives 98491965