The notation is 49C6.
The formula for combination: nCr = n! / {r! * (n-r)!}
Recall that m! = m*(m-1)*(m-2)*...*2*1
That is the official way. You can use a calculator to find n! r! and (n-r)!
We can also use hand-calculation by applying the formula in a different fashion. That is nCr = {(n * (n-1) * ... * (n-r+1) / r!} Let us do hand-calculation to solve the problem.
49C6 = (49 * 48 * 47 * 46 * 45 * 44) / ( 6 * 5 * 4 * 3 * 2 * 1) = 13983816.
=========================
there are 13,983,816 combinations.
15/49ExplanationThere are 15 prime numbers between 1 and 49 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47). If you randomly choose one natural number from the 49 numbers between 1 and 49 inclusive, there is a 15/49 probability that it will be prime.
It is 1/13,983,816.
To find the mean you add all the numbers together then divide by how many there are. Ex. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from least to greatest the find the number in the middle. (note: if there are two numbers in the middle add hem and divide by two.) Ex. 5,5,5,6,6,10,12 Median:6 To find the mode you look for the number that appears the most. (note: There can be more than one mode or there can be no mode) Ex. 5,5,5,6,6,10,12 Mode:5
There are 4 choices for the first digit (1, 2, 3, or 4) since zero cannot be the first digit in a four-digit number. For the second digit, there are 3 remaining choices, for the third digit there are 2 choices, and for the fourth digit, there is only 1 choice left. Therefore, the total number of four-digit numbers that can be formed using the digits 1, 2, 3, and 4 is 4 x 3 x 2 x 1 = 24.
nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.
there are 13,983,816 combinations.
13,983,816
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)
There are 51C6 = 51*50*49*48*47*46/(6*5*4*3*2*1) = 18,009,460 combinations.
if we do not want to use the same number more than once then the answer is: 49*48*47*46*45*44 however if we can use a number more than once the solution is: 49^6
No. There are nearly 14 million combinations of 49 things taken 6 at a time. Excel does not have that many rows or columns to support that.
7
there are many combinations of numbers that add up to 84. Some examples are 41+43, 35+49, 29+55, 24+61, and 18+66.
63