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There are 15 combinations.
3*5=15 5*3=15 15*1=15
The three odd numbers that add up to 43 are 13, 15, and 15. This is because 13 + 15 + 15 = 43. These numbers are odd because they are not divisible by 2, and their sum equals 43, satisfying the condition of the question.
Numbers are unique so that only 15 is equal to 15. However, the limit of 14.999... repeating is 15.
10 000 * * * * * NO! That is the number of PERMUTATIONS, not COMBINATIONS. In a combination, the order does not matter so that 1234 is the same as 1432 or 3412 etc. Assuming the 4 numbers are different, the correct answer is 15 comprising 4 1-digit combinations, 6 2-digit combinations, 4 3-digit combinations and 1 4-digit combination. Another way to look at it is that the first number can be in a combination or not. With each of these possibilities, the second can be in or out - giving 2*2 = 4 ways so far. With each of these there are two options for the third giving 2*2*2 = 8 combinations so far and then the last number makes it 2*2*2*2 = 16. But one of these combinations contains none of the numbers - each one is not in. Leaving that one out gives the answer 15. In general, the number of combinations of any size, from n distinct objects is 2n and if you exclude the null combination, it is 2n - 1.