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3 years ago

11. How many different ways can the 16 numbered pool balls be placed in a line on the pool table?

Mathematics

1 answer:

nikklg [1K]3 years ago

7 0

Answer:

The number of ways are 16! or 20,922,789,888,000.

Step-by-step explanation:

Consider the provided information.

We need to determine the number of different ways 16 numbered pool balls be placed in a line on the pool table.

For the first place we have 16 balls.

For the second place we have 15 balls left.

Similarly for the third place we have 14 balls as two balls are already arranged and so on.

Or we can say that this is the permutation of 16 things taking 16 at a time.

Thus the number of ways are: $16!$ or $^{16}P_{16}$

$16!=16\times15\times14\times13......\times2\times1=20,922,789,888,000$

Hence, the number of ways are 16! or 20,922,789,888,000.

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The n candidates for a job have been ranked 1, 2, 3,..., n. Let X 5 the rank of a randomly selected candidate, so that X has pmf

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Question:

The n candidates for a job have been ranked 1, 2, 3,..., n. Let x = rank of a randomly selected candidate, so that x has pmf:

$p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.$

(this is called the discrete uniform distribution).

Compute E(X) and V(X) using the shortcut formula.

[Hint: The sum of the first n positive integers is $\frac{n(n +1)}{2}$ , whereas the sum of their squares is $\frac{n(n +1)(2n+1)}{6}$

Answer:

$E(x) = \frac{n+1}{2}$

$Var(x) = \frac{n^2 -1}{12}$ or $Var(x) = \frac{(n+1)(n-1)}{12}$

Step-by-step explanation:

Given

PMF

$p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.$

Required

Determine the E(x) and Var(x)

E(x) is calculated as:

$E(x) = \sum \limits^{n}_{i} \ x * p(x)$

This gives:

$E(x) = \sum \limits^{n}_{x=1} \ x * \frac{1}{n}$

$E(x) = \sum \limits^{n}_{x=1} \frac{x}{n}$

$E(x) = \frac{1}{n}\sum \limits^{n}_{x=1} x$

From the hint given:

$\sum \limits^{n}_{x=1} x =\frac{n(n +1)}{2}$

So:

$E(x) = \frac{1}{n} * \frac{n(n+1)}{2}$

$E(x) = \frac{n+1}{2}$

Var(x) is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculating: $E(x^2)$

$E(x^2) = \sum \limits^{n}_{x=1} \ x^2 * \frac{1}{n}$

$E(x^2) = \frac{1}{n}\sum \limits^{n}_{x=1} \ x^2$

Using the hint given:

$\sum \limits^{n}_{x=1} \ x^2 = \frac{n(n +1)(2n+1)}{6}$

So:

$E(x^2) = \frac{1}{n} * \frac{n(n +1)(2n+1)}{6}$

$E(x^2) = \frac{(n +1)(2n+1)}{6}$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = \frac{(n+1)(2n+1)}{6} - (\frac{n+1}{2})^2$

$Var(x) = \frac{(n+1)(2n+1)}{6} - \frac{n^2+2n+1}{4}$

$Var(x) = \frac{2n^2 +n+2n+1}{6} - \frac{n^2+2n+1}{4}$

$Var(x) = \frac{2n^2 +3n+1}{6} - \frac{n^2+2n+1}{4}$

Take LCM

$Var(x) = \frac{4n^2 +6n+2 - 3n^2 - 6n - 3}{12}$

$Var(x) = \frac{4n^2 - 3n^2+6n- 6n +2 - 3}{12}$

$Var(x) = \frac{n^2 -1}{12}$

Apply difference of two squares

$Var(x) = \frac{(n+1)(n-1)}{12}$

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