For English speaking people, they are in alphabetical order.
4+1=5... 5x2=10... 10+3=13 !
Order is not important, so this is a combination problem. There are 5 odd numbers from 1 to 10. n(E) = 5C2 = 5*4/2*1 = 10 n(S) = 10C2 = 10*9/2*1 = 45 P = n(E)/n(S) = 10/45 = 2/9
1 + 9 = 10 2 + 8 = 10 3 + 7 = 10 4 + 6 = 10 5 + 5 = 10 6 + 4 = 10 7 + 3 = 10 8 + 2 = 10 9 + 1 = 10 to continue: 1 + 1 + 8 = 10 2 + 1 + 7 = 10 3 + 1 + 6 = 10 4 + 1 + 5 = 10 5 + 1 + 4 = 10 6 + 1 + 3 = 10 7 + 1 + 2 = 10 8 + 1 + 1 = 10 9 + 1 + 0 = 10 As you can see from the tables above, you could go on all night to find how many ways to add whole numbers in order to equal 10.
The commutative property is that the order of the operands (numbers) does not matter. So:It is true for: Adding numbers: 1 + 2 = 2 + 1 = 3Multiplying numbers: 2 x 3 = 3 x 2 = 6It is false for: Subtracting numbers: 1 - 2 = -1 ≠ 2 - 1 = 1Dividing numbers: 3 ÷ 6 = 1/2 ≠ 6 ÷ 3 = 2
The prime numbers between 1 and 10 are 2, 3, 5, and 7.
This is the sequence of numbers from 1 to 10, which, when written as English words, they would be in alphabetical order.
2
4+1=5... 5x2=10... 10+3=13 !
The prime numbers from 1 to 10 are: 2, 3, 5, 7
Order is not important, so this is a combination problem. There are 5 odd numbers from 1 to 10. n(E) = 5C2 = 5*4/2*1 = 10 n(S) = 10C2 = 10*9/2*1 = 45 P = n(E)/n(S) = 10/45 = 2/9
There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.
1, 2, 5, 10
1 & 10 2 & 5
1 + 9 = 10 2 + 8 = 10 3 + 7 = 10 4 + 6 = 10 5 + 5 = 10 6 + 4 = 10 7 + 3 = 10 8 + 2 = 10 9 + 1 = 10 to continue: 1 + 1 + 8 = 10 2 + 1 + 7 = 10 3 + 1 + 6 = 10 4 + 1 + 5 = 10 5 + 1 + 4 = 10 6 + 1 + 3 = 10 7 + 1 + 2 = 10 8 + 1 + 1 = 10 9 + 1 + 0 = 10 As you can see from the tables above, you could go on all night to find how many ways to add whole numbers in order to equal 10.
There are 2 numbers that go into both 10 and 12. They are 1 and 2.
The commutative property is that the order of the operands (numbers) does not matter. So:It is true for: Adding numbers: 1 + 2 = 2 + 1 = 3Multiplying numbers: 2 x 3 = 3 x 2 = 6It is false for: Subtracting numbers: 1 - 2 = -1 ≠ 2 - 1 = 1Dividing numbers: 3 ÷ 6 = 1/2 ≠ 6 ÷ 3 = 2
The prime numbers between 1 and 10 are 2, 3, 5, and 7.