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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?

how would you go about ans this question?

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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

Therefore, for your question

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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Start your 7-day free trialaabbbcccdd 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

u can added up:

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............

it's been weeks after giving out answer to the question...........where's

my answer is correct

C++

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