# Probability Question

if you have 10 students who can recieve a grade of A,B,C,D or F how many ways can grades be assigned if you know that at least two of them get A's?

how would you go about ans this question?
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Commented:
School assignment. Let's wait for nietod's offical statement.
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Author Commented:
it's a brain teaser that's been bugging me for last 2 weeks....
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Commented:
dont know if this's correct

let take a simple example, at least 2 out of 3 students assigned grade A

ans:
ways assigned with at least 2 A's = Total ways - ways assigned with no grade A - ways assigned with only 1 grade A

Totals ways = 5^3
ways assigned with no grade A = 4^3
ways assigned with 1 grade A = 3 * 4^2

Totals ways = 5^10
ways assigned with no grade A = 4^10
ways assigned with 1 grade A = 10 * 4^9

ans = 5^10 - 4^10 - 10*4^9 = 6095609
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Commented:
does position matter? Is aabbbcccdd
considered the same as ddcccbbbaa for the 10 students?
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Commented:
no  the position doesnt matter.........

aabbbcccdd not equal to ddcccbbaa, considering they are different students standing different position.........

even ababbbcccd not equal ddcccbbaba, they are not the same.........

important note: yr question is at least 2 A, not just 2 A, it includes 3 A , 4 A, 5 A and etc.......

Just imagine 10 different students have their distinct positions to stand...........

5^10 = number of ways to assign grade (5 different grades)   do u agree?

ways assigned with no grade A = 4^10  do u agree?

ways assigned with only 1 A = 10*(4^9)
do u agree?

The formula i used is called binomial equal/theorem........

ways assigned with no A + ..with 1 A + .. with 2 A + .. 3 A +........10 A
the result will be 5^10

the formala listed below
(x+y)^n = C(n,0)(x^0)(y^n) + C(n,1)(x^1)(y^(n-1)) +   ... + C(n,n) (x^n)(y^0)

where C(n,r) = P(n,r)/r! = n!/(r!(n-r)!)
C(n,r) sometimes read as "n choose r"

By using this formula,
n = the number of students = 10
x+y = number of grades = 5 (A,B,C,D,F)

total number of ways = (x+y)^n = 5^n

let x =4 and y = 1, u will get wat u wanted.......

ways with no A = C(n,0)(x^0)(y^n) = 4 (1^0) (4^10) = 4^10

ways with only 1 A = C(n,1) (x^1)(y^9) = 10 (1^1)(4^9)

total ways - ways with no A - ways with 1 A = ways with at least 2 A

or alternatively
ways with 2 A + ways with 3 A + ways with 4 A +........+ ways with 10A
they are the same........
eg. ways with 2 A = C(n,2) (x^2)(y^8) = 45 (1^2) (4^8)

eg. ways with 3 A = C(n,3) (x^3)(y^7) = 120 (1^3) (4^7)

added up til "ways with 10 A"
the result will be the same.......

it's a bit confusing, i am quite sure it's correct............
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Commented:
hmm.....

where's the pts that i supposed to earn..........

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Commented:
hmm.......

it's been weeks after giving out answer to the question...........where's the pts i supposed to get.........