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ale4655 [162]
4 years ago
11

There are 9 students running for 4 openings on the student council. Only students who are running for election can be elected; n

o write-in candidates are allowed. An elected student can only hold one position at a time. Use Pascal’s Triangle to find the number of different ways 4 students can be elected from the group of 9 candidates.
Mathematics
1 answer:
slavikrds [6]4 years ago
6 0
9 students, 4 are selected and order doesn't matter:

Go to the 9th row of the triangle (mind you the first row is 0), the 2nd is 1st row

9th row: 1   9   36   84   126   126   84   36   9   1
Now choose the 4th column (excluding the 1st , you will get 126 :
 so 126 is the number of combination requested:

PROOF: ⁹C₄ = 9!/(9-4)! .4! = (9!)/5!.4! = 126
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Answer:

BD - AC = 0

Step-by-step explanation:

Given that:

BC = 36 , CD = 15 and BD = x

As these are forming right angled triangle,

Using Pythagorean theorem,

(BD)^2+(CD)^2=x^2\\(36)^2+(15)^2 = x^2\\1296 + 225 = x^2\\1521 = x^2\\

Taking square root on both sides

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As the given shape is rectangular, the sides parallel to each other will have same values.

AB = 15 , AD = 36 and AC = x

(15)^2+(36)^2 = x^2 \\225 + 1296 = x^2 \\1521 = x^2 \\x = 39

Now,

BD - AC = 39 - 39 = 0

Hence,

BD - AC = 0

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