probabilities

RauLiTo · 2006-12-26 01:07:34

i have today a confusing question in the probability and its say :

if we have X : { 0 , 1 , 2 , 3 , 4 , 5 }
how many numbers can we make from that group without using the same digits in the same number ... the number contain 4 digits and it must be individual (( doesnt divide by 2 ))

please answer me ... i need help tomorrow is my final exam in school !

Last edited by RauLiTo (2006-12-26 01:13:44)

RauLiTo · 2006-12-26 02:11:26

maybe its not clear ... ?!

mathsyperson · 2006-12-26 02:18:02

I've never seen the definition "individual" as meaning 'not divisible by 2' before. But whatever.

Let's sort out that problem first. If we want a number to not be divisible by 2, then it has to end in 1, 3, 5, 7 or 9. That means that out of your choices, there are 3 that will fit the condition.

The other digits have no restrictions upon them, other than that they can't be repeats of other digits.

So the first digit can be any of the 6 except the one that the last digit was (it has a choice of 5), the second digit has the same choice as the first except for the one that the first one took (it has a choice of 4), and similarly the third digit has a choice of 3.

So combining all of this together, the total amount of combinations is given by 3*5*4*3 = 180.

RauLiTo · 2006-12-26 02:22:34

sorry ... but i dont know what do they call those numbers which doesnt divide by 2 in english !
when i read the question i answer it immediately like you but then i thought about it ...
we cant put '' 0 '' in the last digits because it will be a number with 3 digits ! ...

thanks

mathsyperson · 2006-12-26 02:34:26

Ooh, you're absolutely right. Well spotted. In that case, we need to place another restriction on the first digit.

Now, the first digit can be any of 1, 2, 3, 4 or 5, excluding the one that the last digit was. The last digit can't be 0, so it will have always taken one of these options (which is lucky, because otherwise it would have been more complicated to get an answer). So now the first digit only has 4 options. The second and third digits still have the same amount of choices, and the reasoning for them is unchanged.

So now your answer is 3*4*4*3 = 144.

Sorry for missing that before.

RauLiTo · 2006-12-26 03:01:26

excuse me can you explain it more for me please ?! :$

mathsyperson · 2006-12-26 03:15:24

We've already established that the last digit has to be one of 1, 3, or 5, so that it won't be divisible by 2. That means that it has 3 possibilities.

So that the number is 4 digits long, the first digit is not allowed to be 0. So, depending on what the last digit is, the set of numbers it can be is either {2,3,4,5}, {1,2,4,5} or {1,2,3,4}. Each of these sets have 4 numbers in them, so the first digit will always have 4 choices.

The second digit has no restrictions on it, except that it can't be the same as the first or last digit, because they have already been taken. So, it can have any of the 6 numbers, minus the two that have been taken. So, 4 choices. Similarly, the third digit can have any of the 6 numbers, minus the three that have been taken (3 choices).

And so the total amount of possibilities is 3*4*4*3 = 144.

RauLiTo · 2006-12-26 03:25:09

thanks alot alot mathsy ... you really helped !

Math Is Fun Forum

#1 2006-12-26 01:07:34

probabilities

#2 2006-12-26 02:11:26

Re: probabilities

#3 2006-12-26 02:18:02

Re: probabilities

#4 2006-12-26 02:22:34

Re: probabilities

#5 2006-12-26 02:34:26

Re: probabilities

#6 2006-12-26 03:01:26

Re: probabilities

#7 2006-12-26 03:15:24

Re: probabilities

#8 2006-12-26 03:25:09

Re: probabilities

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