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GenaCL600 [577]
3 years ago
11

A club of twenty students wants to pick a three person subcommittee. How many ways can this be done?

Mathematics
1 answer:
natulia [17]3 years ago
8 0

1140 ways are there to select three person subcommittee from a club of twenty students

<em><u>Solution:</u></em>

Given that a club of twenty students wants to pick a three person subcommittee

To find: number of ways this can be done

We have to use combinations formula to solve the given sum

A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected

<em><u>The formula to calculate combinations is:</u></em>

n C_{r}=\frac{n !}{(n-r) ! r !}

where n represents the number of items, and r represents the number of items being chosen at a time

<em><u>Here we have to choose 3 persons from 20 students</u></em>

So, n = 20 and r = 3

\begin{aligned}&20 C_{3}=\frac{20 !}{(20-3) ! 3 !}\\\\&20 C_{3}=\frac{20 !}{17 ! 3 !}\end{aligned}

<em><u>To get the factorial of a number n ,the given formula is used, </u></em>

n !=n \times(n-1) \times(n-2) \dots \times 2 \times 1

Therefore,

20 C_{3}=\frac{20 \times 19 \times 18 \times 17 \ldots \ldots \ldots 2 \times 1}{17 \times 16 \times 15 \ldots .2 \times 1 \times 3 !}

20 C_{3}=\frac{20 \times 19 \times 18}{3 !}=\frac{20 \times 19 \times 18}{3 \times 2 \times 1}=1140

Thus 1140 ways are there to select three person subcommittee from a club of twenty students

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<em><u>Solution:</u></em>

Given system of equations are:

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We can solve the above system of equations by elimination method

<em><u>Multiply eqn 1 by 4</u></em>

4(-1x + 2y = -4)

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<em><u>Add eqn 2 and eqn 3</u></em>

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<em><u>Substitute y = -1 in eqn 1</u></em>

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<em><u>Check the answer:</u></em>

Substitute x = 2 and y = -1 in eqn 2

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Thus the obtained answer is correct

Thus the solution to given system of equations is (x, y) = (2, -1)

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The value of expanding (2x -3)^4 is 16x^4  + 96x^3  +216x^2 -216x + 81

<h3>How to expand the expression?</h3>

The expression is given as:

(2x -3)^4

Using the binomial expansion, we have:

(2x -3)^4 = ^4C_0 * (2x)^4 * (-3)^0 +^4C_1 * (2x)^3 * (-3)^1 + ^4C_2 * (2x)^2 * (-3)^2 + ^4C_3 * (2x)^1 * (-3)^3 + ^4C_4 * (2x)^0 * (-3)^4

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So, we have:

(2x -3)^4 = 1 * (2x)^4 * (-3)^0 + 4 * (2x)^3 * (-3)^1 + 6 * (2x)^2 * (-3)^2 + 4 * (2x)^1 * (-3)^3 + 1 * (2x)^0 * (-3)^4

Evaluate the exponents and the products

(2x -3)^4 = 16x^4  + 96x^3  +216x^2 -216x + 81

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Read more about binomial expansions at:

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