How do i solve z=a/m, for a
2 answers:
You might be interested in
9514 1404 393
Answer:
300
Step-by-step explanation:
There are 25 ways to select the first student. After that student is removed from the selection pool for the second student, there are 24 ways to select the second student. This gives 25·24 = 600 ways to select 2 students <em>in a particular order</em>.
Since we don't care about the order, we can divide this number by the number of ways two students can be ordered: AB or BA, 2 ways.
600/2 = 300
There are 300 ways to pick a combination of two students from 25.
__
<em>Additional comments</em>
This sort of selection (2 out of 25) has a formula for it, and an abbreviation for the formula.
"n choose k" can be written nCk or C(n, k)
The function is a ratio of factorials:
nCk = n!/(k!(n-k)!)
If you can typeset this, it is written ...
This is different from the formula for the number of <em>permutations</em> of n things taken k at a time. That would be written nPk or P(n, k) = n!/(n-k)!.