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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.