(a)	How many different arrangements can be made with the letter of the word 
‘M O R O C C O’ 
(b)	In how many ways can a team of 5 boys be selected from a group of seven boys?
(c)	A class consists of 6 girls and 10 boys. If a committee of three is chosen at random from the class, find the probability that three boys are selected


Let's tackle each part of your question one by one.

### (a) Different Arrangements of the Word "MOROCCO"

To find the number of different arrangements of the letters in the word "MOROCCO", we use the formula for permutations of multiset:

\[
\text{Total arrangements} = \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}
\]

Where:
- \( n \) is the total number of letters.
- \( n_1, n_2, \ldots, n_k \) are the frequencies of the repeating letters.

For the word "MOROCCO":
- Total letters, \( n = 7 \)
- The letter "O" appears 2 times.
- The letter "C" appears 2 times.

Thus, the number of different arrangements is:

\[
\text{Total arrangements} = \frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260
\]

So, there are **1260 different arrangements** of the letters in "MOROCCO".

### (b) Number of Ways to Select 5 Boys from 7

This is a combination problem where we select 5 boys out of 7 without regard to the order. The formula for combinations is:

\[
\text{Number of ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

Where:
- \( n = 7 \) (total number of boys)
- \( r = 5 \) (number of boys to be selected)

So,

\[
\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21
\]

Thus, there are **21 different ways** to select 5 boys from a group of 7.

### (c) Probability of Selecting 3 Boys from 6 Girls and 10 Boys

The probability of an event is given by:

\[
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
\]

1. **Total Number of Outcomes:**
   - The total number of ways to choose 3 students from 16 (6 girls + 10 boys) is given by \( \binom{16}{3} \):

   \[
   \binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560
   \]

2. **Number of Favorable Outcomes:**
   - The number of ways to choose 3 boys from 10 boys is given by \( \binom{10}{3} \):

   \[
   \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
   \]

Thus, the probability of selecting 3 boys is:

\[
\text{Probability} = \frac{120}{560} = \frac{3}{14} \approx 0.214
\]

So, the probability that three boys are selected is approximately **0.214** or **21.4%**.

---

Would you like further details or have any questions on this? Here are five related questions to consider:

1. How would the arrangements change if another letter was added to "MOROCCO"?
2. What if you had to select 4 boys from a group of 7? How would the calculation change?
3. What if the class had an equal number of girls and boys? How would the probability change?
4. How do permutations differ from combinations in other practical problems?
5. What are some real-life situations where these types of probability problems might arise?

**Tip:** Remember that permutations consider order, while combinations do not. This distinction is crucial when solving problems involving arrangements or selections.

How many different arrangements can be made with the letters of the word 'MOROCCO'? In how many ways can a team of 5 boys be selected from a group of seven boys? A class consists of 6 girls and 10 boys. If a committee of three is chosen at random from the class, find the probability that three boys are selected.

Explore the detailed solutions to permutation, combination, and probability problems, including the arrangement of letters in 'MOROCCO', selecting a team of 5 boys from 7, and calculating the probability of choosing 3 boys from a class of 16 students. Suitable for Grades 9-12.

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