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A pencil case can hold seven pens and eight pencils. Meg chooses either a pen or pencil. Then John is to choose a pen and a pencil. In which case (based on Megs initial choice) would John have to have the greater number of choices.
Does this look right?
If Meg chooses pen:
7C1x6C1x8C1
=336
If Meg chooses pencil:
8C1x7C1x7C1
=392
Therefore, John will have more choices if Meg chooses a pencil.
Hi CroatBoy;
n! / (n -2)! = n * ( n-1) because
n! = n * (n-1) * (n-2) * (n-3) * ... (1)
(n-2)! = (n-2) * (n-3) * ... (1) So just cancel:
Ok. Thanks.
Hi;
Simplifcations are possible :
n! / ( n-2 ) ! = n ( n-1 )
n - (n - 2) = 2
So:
Why does it become 1?
Hi;
Did you understand how to get ( n ( n - 1) ) / 2 = 10 from:
I think so, but could you go over it?
Hi;
Expand it:
Times by 2:
And factor:
You can set both equal to 0, like this
Can you solve these 2 linear equations? Do you need help?
Thanks alot, makes sense now. Appreciate it!
Hi;
Expand it:
Times by 2:
And factor:
Sorry, that is foreign to me. I haven't solved questions like that in the course.
Here's another way:
:
Think of the formula for
Now rewrite using that. You should be able to solve for n.
Tell me if you need any more help.
So it would be n!/(n-2)!(n- (n-1) )! = 10 ?
I'm not sure how to solve this.
Hi CroatBoy;
Solve (n (n-1)) / 2 = 10
You get n = 5 and n = - 4
But how do you get from (n)(n-1)/2 = 10 to 5 and -4. i know it needs to be factored, but im not sure what the steps are.
Solve the equation for n algebraically.
(n)C(n-2) = 10
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