Permutations And Combinations

We are given that there are 8 men and 5 women in a panel, and we are asked to find the number of ways to form a committee of 3 women and 5 men.

The number of ways to choose 3 women from 5 women can be calculated using the combination formula:

Where is the total number of items, and is the number of items to choose.

For choosing 3 women from 5, we have:

Next, the number of ways to choose 5 men from 8 men is:

Now, to form the committee, we multiply the two values together:

The number of ways to form the committee is 560.

We are given that:

The formula for is:

We start with the first equation: .

Using the formula:

We are given that this equals 42, so:

Multiplying both sides by 2:

We now proceed with the second equation: .

We are given that this equals 20, so:

Multiplying both sides by 2:

Now we have the following system of equations:

We solve these equations by trial and error or algebraically.

If we try , we get:

, which is correct.

For , we get:

, which is correct.

Therefore, and .

The values of and are 5 and 2, respectively.

We are asked to find the number of 3-digit numbers that can be formed from the digits 0, 2, 3, 5, and 7, where repetition is allowed.

For a 3-digit number, the first digit cannot be 0, so it must be one of the digits 2, 3, 5, or 7. Therefore, there are 4 possible choices for the first digit.

The second and third digits can be any of the 5 given digits (0, 2, 3, 5, or 7), so there are 5 choices for each of these digits.

Thus, the total number of 3-digit numbers is:

The number of 3-digit numbers that can be formed is 100.

The word ALRIGHT contains 7 distinct letters. If the arrangement must begin with "LG", we treat "LG" as fixed at the beginning of the arrangement. This leaves 5 remaining letters: A, R, I, H, T. The number of ways to arrange these 5 letters is given by the factorial of 5:

Thus, the number of arrangements of the word ALRIGHT that begin with "LG" is .

To form a number greater than 6000, the first digit must be either 6 or 7 (since the number must be greater than 6000).
Therefore, we have two choices for the first digit. Once the first digit is selected, we are left with four digits to choose from for the remaining three places.
Since no digit can be repeated, the total number of ways to arrange the remaining three digits is the product of choices for each place.
For the first digit, we have 2 choices: 6 or 7.
For the second digit, we have 4 remaining digits to choose from.
For the third digit, we have 3 remaining digits to choose from.
For the fourth digit, we have 2 remaining digits to choose from.
Thus, the total number of numbers greater than 6000 that can be formed is:
Since we have 2 choices for the first digit (6 or 7), the total number of numbers greater than 6000 is:

The correct option is (C) :

We are given two formulas:

Step 1: Use the formula for permutation

The formula for permutation is: For , we have: Substitute : Now calculate : Thus, the equation becomes: Now, simplify:

Step 2: Find

Now, check the factorial values to find : So, , hence .

Step 3: Verify using the combination formula

We also have the combination formula: Substitute and : Now calculate: This confirms that .

The correct option is (C) : 4

Given:

condition is

So lets put

Then the parent term will be

So we can represent the term as :

Given: 5 identical dogs, 6 identical monkeys, and 7 identical horses.

We need to find the number of ways to select one or more animals from these groups.

For each type of animal, the number of selection options is:

• Dogs: Can select 0 to 5 (6 options)
• Monkeys: Can select 0 to 6 (7 options)
• Horses: Can select 0 to 7 (8 options)

Total number of combinations (including selecting none):

Since we want at least one animal, we subtract the case where we select none:

Therefore, the number of ways to select one or more animals is 335.

The correct answer is (D) 335.

Given: We need to distribute identical balls into distinct boxes.

The number of ways to distribute identical items into distinct boxes is given by the formula:

This formula comes from the stars and bars theorem, where we represent the balls as stars and the dividers between boxes as bars. We need bars to divide stars into groups.

Let's verify the options:

• (A) - Incorrect, this would be for choosing balls out of distinct balls
• (B) - Incorrect, similar issue as (A)
• (C) - Correct, matches our formula
• (D) - Incorrect, this would be for positive integer solutions
• (E) - Incorrect, this would be if we had additional items

Therefore, the correct answer is (C) .