What is the answer to 2+1/6y=3x+4

Case A = there are 10 zeros and no ones
Case B = there are 9 zeros and 1 one
Case C = there are 8 zeros and 2 ones
Case D = there are 7 zeros and 3 ones
Case E = there are 6 zeros and 4 ones

Any other case isn't possible since we want more 0s than 1s



For case A, we have 10 C 0 = 1 way to do this which is just a string of 0s, which is length 10. The notation 10 C 0 refers to the n C r combination formula. 

For case B, we have 10 C 1 = 10 ways to do this. Simply move the lone 1 to be placed at each slot giving 10 different combinations.

For case C, we have 45 different combos because 10 C 2 = 45. 

Case D has 120 different outcomes (note how 10 C 3 = 120)

Case E has 210 different outcomes (10 C 4 = 210)

As a shortcut, you can look at Pascal's triangle. Look at the row that starts with 1, 10, 45, ... and record the first 5 values. Those values match up with the results we got.



Now add up those results to get: 1+10+45+120+210 = 386

Therefore, the final answer is 386. There are 386 different bit strings of length 10 that has more 0s than 1s.