Test: Combinations - GMAT MCQ

To calculate the number of clothing combinations Barbara has, we can subtract the number of prohibited combinations from the total number of possible combinations.

Barbara has 8 shirts and 9 pants, resulting in a total of 8 x 9 = 72 possible combinations if she could wear any shirt with any pants.

However, she cannot wear 2 specific shirts with 3 specific pants. This means we need to subtract the number of combinations that include these prohibited combinations.

The number of prohibited combinations is 2 (specific shirts) x 3 (specific pants) = 6.

Therefore, the total number of clothing combinations Barbara has is 72 - 6 = 66.

Hence, the correct answer is B: 66.

In a group of 8 people that includes George and Nina, we need to select 3 people randomly for a project. The objective is to choose a group that includes George but excludes Nina.

To meet this requirement, we consider that George must be included in the selection. Therefore, we only need to choose 2 additional people from the remaining 6 individuals (excluding Nina).

The number of ways to choose 2 people from a group of 6 can be calculated as 6C2, which is equal to 15.

The total number of ways to select 3 people from a group of 8 can be calculated as 8C3, which equals 56.

Hence, the probability of selecting the desired group, which includes George but excludes Nina, is 15/56.

Therefore, the answer is C: 15/56.

The Chief of Naval Operations (N) and the Chief of the National Guard Bureau (G) need to sit next to each other in a seating arrangement. We can treat the pair [N - G] as a single entity.

The seating arrangement can be represented as:

[N - G] - A - B - C - D - E

Since we have a circular arrangement, we can consider [N - G] as a single unit and arrange it with the other individuals (A, B, C, D, E) in (6 - 1)! = 5! = 120 ways.

Additionally, we can arrange [G - N] as a single unit, which gives us another 120 ways.

Therefore, the total number of ways for the two chiefs to sit next to each other is 120 + 120 = 240.

The total number of ways for all seating arrangements, considering the circular arrangement, is (7 - 1)! = 6! = 720.

To determine the number of ways the Chief of Naval Operations does not sit next to the Chief of the National Guard Bureau, we subtract the number of favorable outcomes (240) from the total number of possible outcomes (720):

Number of ways the Chiefs do not sit next to each other = 720 - 240 = 480.

Hence, the Chiefs have 480 ways of not sitting next to each other.

To find the number of different possible lineups for Team A in the game of tug-of-war, we need to consider the arrangements of the males and females in the lineup.

Since there are 3 males and 3 females, we can arrange them within their respective groups in 3! (3 factorial) ways for each group. Therefore, the number of ways to arrange the males and females in the lineup is:

3! (for males) × 3! (for females) = 6 × 6 = 36

Hence, the correct answer is D: 36 different possible lineups.

In the given word "GMATINSIGHT," there are 11 letters. The letter 'T' appears twice, the letter 'I' appears twice, and the remaining letters 'G,' 'M,' 'A,' 'N,' 'S,' and 'H' each appear once.

Since the first letter 'G' and the last letter 'T' are fixed, we need to arrange the remaining 9 letters. However, since the letter 'I' appears twice, we need to consider the arrangements where the two 'I's are swapped as identical.

Therefore, the total number of desired outcomes can be calculated as follows:

Total arrangements of 9 letters = 9!

Arrangements repeated due to swapping of identical letter 'I' = 2!

Hence, the total desired outcomes equal 9! divided by 2!.

There are a total of 10 possible combinations for packages that include the same flavor of tea (2 x 5 = 10).

Considering the variety pack with 5 flavors, there are 5 choices possible when selecting 4 flavors (5C4 = 5). Since each of these choices can be presented in 2 forms, there are 10 possibilities for packages that contain different flavors (2 x 5 = 10).

In summary, the total number of different package types is 20, which is obtained by adding the 10 packages with the same flavor to the 10 packages with different flavors (10 + 10 = 20).

Let's represent the desks as D1, D2, D3, and D4. The students can be seated in the following arrangements:

• Empty Desk - Student - Empty Desk - Student (E-S-E-S): There are two possible ways to arrange the students in this configuration: Student 1 in D1 and Student 2 in D3 or vice versa.
• Student - Empty Desk - Empty Desk - Student (S-E-E-S): There are two possible ways to arrange the students in this configuration: Student 1 in D1 and Student 2 in D4 or vice versa.
• Empty Desk - Student - Student - Empty Desk (E-S-S-E): There are two possible ways to arrange the students in this configuration: Student 1 in D1 and Student 2 in D4 or vice versa.

Combining these possibilities, we have a total of 2 + 2 + 2 = 6 different ways to seat the students with at least one empty desk between them.

Therefore, the correct answer is D: 6.

In division A, there are 5 contenders, and we need to determine the number of ways to choose the top 2 positions. Since the order matters (1st and 2nd place), we can use permutations to calculate this. The number of permutations of 5 items taken 2 at a time is denoted as 5P2 and can be calculated as:

5P2 = 5! / (5 - 2)! = 5! / 3! = (5 x 4) / (2 x 1) = 20.

Similarly, in division B, there are also 5 contenders, and we need to calculate the number of ways to choose the top 2 positions. Using the same logic as before, the number of permutations of 5 items taken 2 at a time is:

5P2 = 5! / (5 - 2)! = 5! / 3! = (5 x 4) / (2 x 1) = 20.

Since the two divisions are independent of each other, we can multiply the number of possibilities in each division to find the total number of ways for the contenders to place 1st and 2nd in both divisions:

20 x 20 = 400.

Therefore, the correct answer is E: 400.

To begin with, let's examine the restricted elements, namely children A, B, and G, who must be seated together in a consecutive arrangement of three seats. How many possible arrangements are there for these "three in a row" seats in a row of seven seats?

X X X _ _ _ _

_ X X X _ _ _

_ _ X X X _ _

_ _ _ X X X _

_ _ _ _ X X X

There are a total of five distinct locations where these three children can be seated in a consecutive manner. Now, for any specific group of three seats, we know that child A must occupy the middle seat, allowing for only two possible seating orders: B-A-G or G-A-B. Consequently, the total number of configurations for these three children is 5 multiplied by 2, resulting in 10.

Moving on to the non-restricted elements, which consist of the remaining four children. Once A, B, and G are seated, the remaining four children can be arranged in any order among the four remaining seats. This corresponds to a permutation of the four items, denoted as 4P4, which equals 4! and results in 24 possible arrangements. For each individual seating configuration of A, B, and G, there are 24 distinct ways in which the other children can be seated in the remaining seats.

Finally, let's combine the information using the Fundamental Counting Principle. We have 10 possible arrangements for the first three children and 24 possible arrangements for the remaining four. Multiplying these two numbers together gives us a total number of configurations of 24 multiplied by 10, which equals 240.

Therefore, the answer is 240.