In an 8 bit binary number, what is the total number of combinations of the eight bits?
494551
256 - the highest number that can be expressed is 255 (calculation shown below), but 0 is also a valid combination:\n\n1 1 1 1 1 1 1 1\n\n=> 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128\n\n= 255\n\nThis is in a sequence of powers of 2 = 2^0 + 2^1 +...+ 2^7
698756
First conversion the numbers to binary, then make the OR operation.\n\nExp.:\n8 OR 3.\n\nis: 1000 OR 11\n\n= 1011.\n\nWhich mean: 11 (Dec.)\n\nYou can use the Windows Calc.
566249
1:9, since there are 18 combinations of numbers, and only two add up to 5, 1+4 & 2+3.
1181393
Binary digits stand for powers of 2.  For example, 2^0 is 1, and 1 in binary is 1.  2 is equal to 2^1, so 2 is 10 in binary.  3 is 2^1 + 2^0, so it's 11 in binary.  \n\nSo the joke is really saying there are 2 types of people.  Now you fall into the former.  :-)\n\n*** There are two types of people, those who know binary and those who don't, now you fall into the former category of those who know.
747867
352 = 11 times 32.\n\n11 is 1011 in binary (8+2+1).\n\n32 is 2^5, so I add 5 zeros.  352 is 101100000.
88776
Are you looking for a set of 8 consecutive whole numbers that when you add 7 of them together in different combinations you receive those 3-digit responses?\n\nWell you've only listed 6 possibilities and only one of those is odd.\n\nStarting at 29 the lowest combination is 224 and the highest is 231.  Starting at 28 the lowest is 217.
415435
nope... 01 (in binary) = 1 (in decimal)\n\nand \n\n10 (on binary) = 2 (in decimal)
842276
1,2,3,4,6,8,12,16,24,48 \n\nlolz hope this helps :)
705107
There are 10 kinds of people who understand binary.\n\nThose who do and those who don't.
359204
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers
1096868
If I did the math right there are ten numbers 0,1,2,3,4,5,6,7,8,9\n\nThey must be put in a group of four. Thats 10 to the fourth power. So thats about ........  10,000 possible combonations. If your asking the possible combonation to a lock....... Well good luck unloking it.
51229
9 x 8 = 72 and pi (this is the correct spelling) out to 512 digits is:\n\n3.1415926535897932384626433832795028841971693993751058209\n749445923078164062862089986280348253421170679821480865132\n823066470938446095505822317253594081284811174502841027019\n385211055596446229489549303819644288109756659334461284756\n482337867831652712019091456485669234603486104543266482133\n936072602491412737245870066063155881748815209209628292540\n917153643678925903600113305305488204665213841469519415116\n094330572703657595919530921861173819326117931051185480744\n623799627495673518857527248912279381830119491298336733624
409800
Generaly, becouse bit numbers are on a boundy of an exponet of 2, we can calculate like this:\n\n1024 bits in 1 kilobit\n1024 kilobits in a megabit\nthus: 1024 * 1024 = 1,048,576 bits
289381
The answer to the three switch problem is 8 choices, not 7. The answer figured numerically is 2^3.\n\nWith 8 switches there are 2^8 or 256 combinations possible.\n\nThere is a formula based on the length of the femur:\nhttp://ublib.buffalo.edu/libraries/projects/cases/bones/activity4.html
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
110013
Well, all numbers would have to be represented equally, right?  If there are 27 combination I assume that 1, 1, 1 is included in these combintations, not just every combination that includes all 3 numbers...  However, if all three numbers are represented equally, each number should appear 27 times.  So the answer should be 27*9, or 243.
837543
If I have interpreted you correctly, each of the 49 packages must contain either BBB, BBS, BSS, or SSS.\nThen there are:\n22,100 combinations.\n\nThis is the same as the number of solutions of the equation:\nx1 + x2 + x3 + x4 = 49\nin non-negative integers.\n(i.e., x1 is the number of BBB packages, x2 is the number of BBS, x3 the number of BSS, and x4 the number of SSS)\n\nThis is C(49+4-1, 4-1) = C(52,3) = 52*51*50/(3*2*1) = 22,100.\n\n*************************\nADDITIONAL DETAILS:\nOk, I saw that you reposted this question.  However, you are still missing something or have the numbers or the question incorrect.\n\n285 = 3*5*19 .  This number is not a result formable by the combination function: C(a,b) = a!/[(a-b)!b!]\n\nThe closest would be 286 = C(13,3).  This would be the number of ways that 3 items could be chosen from 13 items.\n\nAlso, as you have it written on the other post, my answer above is still correct.  There are 22,100 ways that 4 different types of items could be distributed into 49 packages.  This assumes that the packages are indistinguishable from each other.  If they weren't, i.e., if BBB in package #1 and BBS in package #2 is different from BBS in #1 and BBB in #2, then the number of combinations is MUCH larger.  (It would be 4^49 ~ 3.2 * 10^29)