combinations of 55If there is not repetition in numbers used......55 x 54 x 53 x 52 x 51 * * * * *No. That is the number of permutations. For combinations the order of the number does not matter so that 1,2,3,4,5 is the same as 1,3,2,4,5 or 1,4,2,3,5 etc.There are 5*4*3*2*1 = 120 such orderings for every set of 5 numbers. So the correct answer is55*54*53*52*51 / 120 = 3478761
The sum you require is 4!/2!2!, which is 6. This means there are 6 different two-digit numbers. * * * * * Wrong! The above formula is for the number of combinations, not permutations. However, since the number 23 is different from the number 32, you require permutations and not combinations. There are 4*3 = 12 23, 25, 27, 32, 35, 37, 52, 53, 57, 72, 73, 75.
The median of a set of numbers is the middle number when they are laid out in numerical order. When there is an even amount of numbers in the sequence like this one, there are two numbers in the middle. To find out the median from these two numbers, you just need to find the average of the two numbers. As the two numbers in the middle of this set of numbers is 53 and 54, the average is 53.5. This means that the median of this set of numbers is 53.5.
There are an unlimited number of possibilities. If you need five consecutive whole numbers, that's different, your answer would b e 51, 52, 53, 54, and 55.
To any set that contains it! It belongs to {-8}, or {-8, sqrt(2), pi, -3/7, 99.3}, or all whole numbers between -43 and 53, or multiples of 2, or integers, or rational numbers, or real numbers, or complex numbers, or square roots of 64 or fourth roots of 4096 etc.
just intrested in the number combinations * * * * * Number of combinations = 56C6 = 56*55*54*53*52*51/(6*5*4*3*2*1) = 32,468,436
There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.
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)
combinations of 55If there is not repetition in numbers used......55 x 54 x 53 x 52 x 51 * * * * *No. That is the number of permutations. For combinations the order of the number does not matter so that 1,2,3,4,5 is the same as 1,3,2,4,5 or 1,4,2,3,5 etc.There are 5*4*3*2*1 = 120 such orderings for every set of 5 numbers. So the correct answer is55*54*53*52*51 / 120 = 3478761
There are 53*52*51*50*49*48/(6*5*4*3*2*1) = 22,957,480 combinations. And you must be joking if you think I am going to list them!
The general formula is to square the number of digits it could be (in this case 5) by how many spaces there are. So in this case it would be 53 = 125 possible combinations.
53 is a prime number
I'm not familiar with "Lotto 556", but if it's meaning is the same as "Lotto 649" which I am familiar with, then it means that five unique numbers between 1 and 56 are picked. If that is correct, then the number of winning combinations are: 56! / 51! Which equals: 56 * 55 * 54 * 53 * 52 or: 458377920
Itself because 53 is a prime number
The prime factors of 53 are 1 and 53, as 53 is a prime number itself. The prime factors of 74 are 1, 2, 37, and 74. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves.
Oh, isn't that a lovely question! To find the number of combinations, we use a formula called combinations. So, for 6 numbers out of 56, the number of combinations would be 56 choose 6, which equals 32,468,436. Isn't that just a happy little number? Just remember, math is like painting a beautiful landscape - it's all about finding the right colors and creating something wonderful.
They are: 1*53 = 53 because it is a prime number