16*15*14*13/(4*3*2*1) = 1820
Since there are 6 numbers that equal 26 to choose from, there are 6 to choose from for the first number, 5 to choose from for the second number, 4 to choose from for the third number, and so on. Therefore there are 6*5*4*3*2*1 ways or 6! or 720.
There are 4 numbers to choose from for the first space, 3 for the next space, and so on. Therefore the answer is 4! or 24.
Assuming no repetition is allowed, there are 8582777280 ways in which you can pick any of the numbers from 1 to 99 inclusive with 5 numbers. This is given by the formula n! / (n - r)! where n is the number of numbers you have to choose from, and r is the number of numbers chosen.
There are ten digits 0-9 that you can choose from. 10P4 = is the number of ways to order these ten digits into four digit numbers.However, 0 cannot begin the number or else it is a 3 digit number (0341 is a three digit number). So, we must subtract from 5040 the number of 3 digit numbers which do not contain 0. There are 1-9 digits to choose from. 9P3 = 504 is the number of ways to order these nine digits into three digit numbers.5040 - 504 = 4536______________________________________________Another way to arrive at the same result:You have 9 choices for the first number (1,2,3,...,9). Once you choose the first number you have 9 numbers to choose from for each of the remaining three numbers: (0,1,2,3,...9) but excluding the number you chose for the first number. So you have (9P3) ways to choose the final three digits.Thus, you have a total of 9*(9P3) ways to form your four digit number:9*(9P3)=9*9*8*7=4536
There is an infinite number of ways three numbers can have the sum of 11
You can choose 3 objects from 6 20 ways, assuming order does not matter
Infinitely many ways - depending on what operations and what set of numbers you choose to allow.
There are an infinite number of ways to do that.
Since it must be a 5 digit number, range is the numbers from 20000 to 55555 we will use multiplication rule to figure this out too. If a number is divible by two, its last(ones place) digit is even. If it is divisible by 5 its last(ones place) digit is zero or 5. We don't give a rat's derriere what the other digits are (for divisibility by two or five) The number of ways to choose the ten thousands place digit is 4 the number of ways to choose the thousands place digit is 5 the number of ways to choose the hundreds place digit is 5 the number of ways to choose the tens place digit is 5 The number of ways to choose the ones place digit is FOUR, since it must be 0,2,4,5 and not 3 use multiplication rule and solution is 4 x 5 x 5 x 5 x 4 = 2000
You can make 125 different 3-digit numbers if you're allowed to repeat a digit (for example, 113, or 545), and 60 if you're not. If repetitions are allowed, then there are 5 ways to choose each digit to be 1,2,3,4, or 5, so the total number of ways is 5*5*5 or 125. If no repetitions are allowed, there are 5 ways to choose the first digit, but only 4 ways left to choose the second (for a total of 5*4=20 ways to choose the first two digits), because it can't be the same as the first. Then you have eliminated two choices for the third digit, so there are only 3 ways to choose it, for a total of 5*4*3=60 different 3-digit numbers.
There are a few ways to do this, but one of the most common ways is by giving each of your subjects a number. If there are 100 of them, number them 001-100. If there are 1000, number them 0001-1000 and so on. Then, randomly choose numbers for each group using random number generator.
12 ways, including those with rows and column numbers swapped.