how many possible combinations of 1 2 3 4?
151182
If you include 3, 2 and 1 digit numbers then it would be 24 + 24 + 12 + 4 = 64 combinations.
16574
Thus you have 10x10x10 = 1000 choices for the first three digits. Finally you have 10 choices for the fourth digit and thus there are 10x10x10x10 = 10 000 possible 4 digit combinations from 0-9.
275875
-- we're in a position to answer your original question. With 8 items on your menu, the number of possible combinations will be 28 = 256. And if you bought all 8 items, and tried to arrange them in different orders on your tray, the number of arrangements ('permutations') would be 8! = 40320.
1092073
By dividing a binary number up into groups of 4 bits, each group or set of 4 digits can now have a possible value of between “0000” (0) and “1111” ( 8+4+2+1 = 15 ) giving a total of 16 different number combinations from 0 to 15.
212052
There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6. 720 / 6 = 120.
38270
Thus you have made 5 × 4 × 3 × 2 1 = 120 choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 4 and 5 if you don't allow any digit to be repeated. Now consider the possibilities with 13 as the first two digits.
114193
reto wrote: How many different 4 letter combinations can be created from the letters A,B,C, and D? This is a case of number of arrangements possible with repetition. Thus you have 4 options each for the 1st,2nd,3rd and 4th letter giving you a total of 4*4*4*4 = 256 number of possible combinations.
414282
Well you have n choices for the nth digit. In your case you have 4 digits. So 4x3x2x1=24 combinations of all 4 digits with none repeating. Without repeating digits 24 four digit numbers can be made .
176805
Where 1,2,3 are the possible options and 3 is the number of repetitions. This will give an array of the 27 different combinations you can achieve.
47858
There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.
966160
Hence, in this case there are 4 3 = 12 possible two digit numbers. For the third digit there are only 2 choices and for the last digit, one choice. Thus if you are not allowed to repeat a digit the number of possible 4 digit numbers you can construct from 1,2,3,4 is 4 3 2 1 = 24.
1031102
Answer and Explanation: The number of possible 16-digit combinations, where repetition is allowed is 10,000,000,000,000,000.
96505
Hence there are 2 choices for the fourth digit and only 1 choice for the fifth digit. Thus you have made 5 × 4 × 3 × 2 1 = 120 choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 4 and 5 if you don't allow any digit to be repeated. Now consider the possibilities with 13 as the first two digits.
225114
There are two possible values for each switch. As there are 4 switches , the total number of permutations or unique combinations are 2^4=16.
1106697
Explanation: There are 5 choices for the first digit. giving 5×5=25 possible combinations. giving 5×4=20 possible combinations.
1370715
In this example there are 2600 possible triplet combinations (26*25*24)/(3*2*1) == 2600 and using the "brute-force" method above, all triplet combinations can be represented in around 301 6-value groups. However, it feels like there ought to be a more efficient way of achieving this.
84525
Hence there are 2 choices for the fourth digit and only 1 choice for the fifth digit. Thus you have made 5 × 4 × 3 × 2 1 = 120 choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 4 and 5 if you don't allow any digit to be repeated.