Thread: what on earth is the formula?

hi im kate, and i need help. we are playing this game (Luffa Lotto) which the no.s are 1,2,3,4,5,6,7,8,9,10. a combination is 1,6,3. another is 9,5,3. 4,6,4 can not be one 'cause there are two fours. we have got to work out how many combinations there are altogther, and we need the formula to work it out. if you know please help me!
thanks soooooooooooooooooooooooooooooooooooooo much
bye

Originally Posted by kate

hi im kate, and i need help. we are playing this game (Luffa Lotto) which the no.s are 1,2,3,4,5,6,7,8,9,10. a combination is 1,6,3. another is 9,5,3. 4,6,4 can not be one 'cause there are two fours. we have got to work out how many combinations there are altogther, and we need the formula to work it out. if you know please help me!
thanks soooooooooooooooooooooooooooooooooooooo much
bye

The best way to do this is try finding the combinations for all numbers starting in one...
1,(2,3,4,5,6,7,8,9,10),(2,3,4,5,6,7,8,9,10)

The second digit can only have nine different numbers and the third can only have eight (because it can't have the same as the second or first one). Therefore the formula for all combinations starting with one is
now we multiply that by 10 to get all valid combinations for the first digit.

Welcome aboard, Kate!

We are playing this game (Luffa Lotto) which the numbers are 1,2,3,4,5,6,7,8,9,10.
A combination is {1,6,3}; another is {9,5,3}.
{4,6,4} cannot be one 'cause there are two fours.
We have got to work out how many combinations there are altogther,
and we need the formula to work it out.

From the examples you gave, the order of the numbers is important.
. . [Technically, you should not call them "combinations".]

You can reason it out like this . . .

For the first number, there are possible choices.

For the second number, there are possible remaining choices.

For the third number, there are possible remaining choices.


Therefore, for selecting three numbers,

. . there are: possible "combinations".

Originally Posted by Soroban

Welcome aboard, Kate!

[size=3]
From the examples you gave, the order of the numbers is important.
. . [Technically, you should not call them "combinations".]

Tsk, Tsk. Picky, picky, picky

Originally Posted by Quick

The best way to do this is try finding the combinations for all numbers starting in one...
1,(2,3,4,5,6,7,8,9,10),(2,3,4,5,6,7,8,9,10)

The second digit can only have nine different numbers and the third can only have eight (because it can't have the same as the second or first one). Therefore the formula for all combinations starting with one is
now we multiply that by 10 to get all valid combinations for the first digit.

yeah but is 720 just the amount of "combinations" for the "combinations" begining with 1?

Hello again, Kate!

yeah but is 720 just the amount of "combinations" for the "combinations" begining with 1?

No, read his explanation again . . .

Therefore the formula for all combinations starting with one is .

Now we multiply that by 10 to get all valid combinations for the first digit.
. .

By the way, did my explanation make sense to you?

Originally Posted by kate

yeah but is 720 just the amount of "combinations" for the "combinations" begining with 1?

No, 720 is the amount of "combinations" without repeating numbers for all possible numbers, not just the ones starting with 1.

Kate do not smiley abuse your post!
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