Question

In how many ways can 7 people be seated around a circular table if three particular people must sit next to each other?

_________

& Task B. Solve for the probability of the given events. Show your complete solution
na hospital, there are 9 nurses and 4 physicians. Five nurses and one physician are males. It
person is selected from the staff, what is the probability that the person is a physician or
ale?​


___________
A. evaluate each expression.

1. C (22, 20) =
2. C (12, 8) =​​

Answer 1

✒️PERMUTATION

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$\large\underline{\mathbb{ANSWER}:}$

$\qquad \LARGE \:\: \rm 144 \: ways$

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$\large\underline{\mathbb{SOLUTION}:}$

Since they are to be arranged in a circular table, we will be using the concept of circular permutations.

$\begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm P = (n-1)!} \end{align}$

If the three wants to be together, then there will be 5 different units to be arranged including the three in which they are considered as one.

• $\rm P = (5-1)!$

• $\rm P = 4!$

• $\rm P = 4 \cdot 3 \cdot 2$

• $\rm P = 24$

By applying the Fundamental Counting Principle, since the three people insisting to sit next to each can swap places, they can be arranged in 3! ways.

• $\rm FCP = 24 \cdot 3!$

• $\rm FCP = 24 \cdot 6$

• $\rm FCP = 144$

Therefore, there are 144 ways to arrange 7 people in the circular table if three particular people must sit next to each other.

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✒️PROBABILITY

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$\large\underline{\mathbb{ANSWER}:}$

$\qquad \Large \:\: \rm Probability \; is \; \frac{9}{13}$

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$\large\underline{\mathbb{SOLUTION}:}$

Let A be the number physicians and B be the number of males. There will be a total of 13 personnel in the hospital.

• $\rm n(S) = 13$
• $\rm n(A) = 4$
• $\rm n(B) = 6$

There is an intersection between A and B since there is a one male physician.

• $\rm n(A\cap B) = 1$

Determine the probability of selecting a physician (A) 'or' a male (B).

• $\small \rm P(A\cup B) = P(A) + P(B) - P(A\cap B)$

• $\small \rm P(A\cup B) = \frac{n(A)}{n(S)} + \frac{n(B)}{n(S)} - \frac{n(A\cap B)}{n(S)}$

• $\small \rm P(A\cup B) = \frac{4}{13} + \frac{6}{13} - \frac{1}{13}$

• $\small \rm P(A\cup B) = \frac{\,10\,}{13} - \frac{1}{13}$

• $\small \rm P(A\cup B) = \frac{9}{13}$

Therefore, the probability of selecting a physician or a male is 9/13

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✒️COMBINATIONS

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$\large\underline{\mathbb{ANSWER}:}$

$\qquad \Large \:\: \rm 1) \; _{22}C_{20}= 231$

$\qquad \Large \:\: \rm 2) \; _{12}C_8 = 495$

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$\large\underline{\mathbb{SOLUTION}:}$

Use the given formula to determine the combinations of n objects taken r at a time.

$\begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm _nC_r = \frac{n!}{r!(n-r)!}} \end{align}$

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Number 1:

Determine the number of combinations of 22 objects taken 20 at a time.

• $\rm _{22}C_{20} = \frac{22!}{20!(22-20)!}$

• $\rm _{22}C_{20} = \frac{22!}{20!\,2!}$

• $\rm _{22}C_{20} = \frac{22 \cdot 21 \cdot \cancel{20!}}{\cancel{20!} \cdot 2}$

• $\rm _{22}C_{20} = \frac{\cancel{22}^{\;\,11} \cdot 21}{\cancel{\,2\,}}$

• $\rm _{22}C_{20} = 11 \cdot 21$

• $\rm _{22}C_{20} = 231$

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Number 2:

Determine the number of combinations of 12 objects taken 8 at a time.

• $\rm _{12}C_8 = \frac{12!}{8!(12-8)!}$

• $\rm _{12}C_8 = \frac{12!}{8!\,4!}$

• $\rm _{12}C_8 = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot \cancel{8!}}{\cancel{8!}\cdot 4 \cdot 3 \cdot 2}$

• $\rm _{12}C_8 = \frac{\cancel{12} \cdot 11 \cdot \cancel{10}^{\;\,5} \cdot 9}{\cancel{\,4\,} \cdot \cancel{\,3\,} \cdot \cancel{\,2\,}}$

• $\rm _{12}C_8 = 11 \cdot 5 \cdot 9$

• $\rm _{12}C_8 = 495$

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*Please read and understand my solution. Don't just rely on my direct answer^^