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  2. What is the formula to get 4 digit combinations from numbers 0-9?

    socratic.org/.../what-is-the-formula-to-get-4-digit-combinations-from-numbers-0-9

    In that case, the number of 4 -digit combinations is given by. 10 ⋅ 9 ⋅ 8 ⋅ 7 = 5,040. Answer link. Please read the explanation, because the answer is either 10,000 or 5,040. Since there are 10 choices for each digit, the number of 4- digit combinations is given by 10*10*10*10 = 10,000, UNLESS using a digit means it cannot be used again.

  3. How do you figure out the number of combinations in 4 digit ... -...

    socratic.org/questions/how-do-you-figure-out-the-number-of-combinations-in-4...

    24" combinations" >"the possible combinations are" "using the 4 digits 1234" ((1,2,3,4),(1,2,4,3),(1,3,2,4),(1,3,2,4),(1,3,4,2),(1,4,2,3),(1,4,3,2))=6((2,1,3,4),(2,1 ...

  4. If we have the numbers 1, 2, 3, and 4, how many combinations ......

    socratic.org/questions/587cdabb11ef6b72cd22cd57

    If we let numbers repeat = 256. If we don't let numbers repeat =24. If we're talking strictly about combinations (vs permutations) = 1. If we are looking at the number of numbers we can create using the numbers 1, 2, 3, and 4, we can calculate that the following way: for each digit (thousands, hundreds, tens, ones), we have 4 choices of numbers. And so we can create 4xx4xx4xx4=4^4=256 numbers ...

  5. How many combinations can you make with the numbers 1,2,3?

    socratic.org/questions/how-many-combinations-can-you-make-with-the-numbers-1-2-3

    Starting with 1 2 3 we can form combinations of size 1 2 or 3. For n things choosing r combinations we can count using the formula. n! r!(n − r)! So we have: 3 choose 1 in 3! 1!(3 −1)! = 3 ways. 3 choose 2 in 3! 2!(3 −2)! = 3 ways. 3 choose 3 in 3! 3!(3 −3)! = 1 way. That is a total of 7 combinations. (If we wish to count choosing 0 ...

  6. How do you calculate combinations of 10 numbers? - Socratic

    socratic.org/questions/how-do-you-calculate-combinations-of-10-numbers

    1 Answer. Wataru. Nov 27, 2014. If r numbers are chosen, then the number of all combinations can be found by. C(10,r) = P (10,r) r! = 10 ⋅ 9 ⋅ ⋯ ⋅ (10 −r +1) r! Example. If r = 3, then. C(10,3) = 10⋅ 9 ⋅ 8 3 ⋅ 2 ⋅ 1 = 120 combinations. I hope that this was helpful.

  7. How many 6 digit combinations can i get using numbers 1-49?

    socratic.org/questions/how-many-6-digit-combinations-can-i-get-using-numbers-1-49

    No. of possible combinations without repeating any number =43,084 Four combinations 6 digits are 1) All ...

  8. A license plate composed of 3 letters followed by 4 digits ... -...

    socratic.org/questions/a-license-plate-composed-of-3-letters-followed-by-4...

    The second letter spot can be any one of the remaining 25 letters. And the third letter spot can be any one of the remaining 24 letters. This gives, for letters: 26xx25xx24=15600 (note - this is the same as a permutation with population 26 and picking 3) We still have 10^4 for the numbers, and so in total we have: 15600xx10000=156,000,000 plates

  9. What are Combinations? - Socratic

    socratic.org/questions/what-are-combinations

    A Combination is a type of calculation that tells us how many ways we can gather together a set of things without regard to the order in gathering those things. A common question in combinations is to calculate how many poker hands are possible (the number of ways we can gather together a certain type of cards, such as a Royal Flush or a Pair ...

  10. Combinations and Permutations - Statistics - Socratic

    socratic.org/statistics/probability/combinations-and-permutations

    The astrological configuration of a party with n guests is a list of twelve numbers that records the number of guests with each zodiac sign. How many different astrological configurations are possible for n = 100? There are 20 rabbits, 15 black and 5 white. You take one by one a rabbit.

  11. How many license plates can be made consisting of 2 letters ... -...

    socratic.org/questions/how-many-license-plates-can-be-made-consisting-of-2...

    The number of different combinations of 2 letters is: 26 xx 26 = 676 The same applies for the three digits. There are 10 choices for the first, 10 for the second and 10 for the third: 10xx10xx10 =1000 So for a license plate which has 2 letters and 3 digits, there are: 26xx26xx10xx10xx10= 676,000 possibilities.