Test: Combinations - GMAT MCQ

Problems that ask us to arrange objects under specific constraints can often be split into smaller, simpler problems.

We'll look for such a division, a Logical approach.

We'll split it into cases:

If we form a number without 2 and without 5, then we must use the number 1,6,8,9.

This gives 4! = 24 options.

If we pick only one of 2,5 and another 3 of the remaining numbers this gives 2C1 * 4C3 = 8 options.

This gives 8 * 4! = 192 arrangements (all of which are good as 2,5 are not together)

If we pick 2, 5, and another 2 of the remaining numbers this gives 2C2 * 4C2 = 6 options.

How many 'bad' options, that is options such that 2 and 5 are together, are there?

3! * 2: 2-5-x-y, 2-5-y-x, x-2-5-y, y-2-5-x, x-y-2-5, y-x-2-5 and all of these with the numbers 2,5 reversed.

This is essentially arranging 3 objects: x,y, and a '2-5' clump.

So - we have (4! - 3!*2) = 12 arrangements per option times 6 options giving 72 arrangements.

24 + 192 + 72 = 288

To determine the number of keystrokes needed to type numbers from 1 to 1000, we can analyze the pattern and calculate it.

From 1 to 9, there are nine single-digit numbers, which require nine keystrokes.

From 10 to 99, there are 90 two-digit numbers, which require 180 keystrokes. Each two-digit number consists of two keystrokes (one for each digit).

From 100 to 999, there are 900 three-digit numbers, which require 2700 keystrokes. Each three-digit number consists of three keystrokes (one for each digit).

Finally, we have the number 1000, which requires four keystrokes.

Adding up the keystrokes for each range, we have:

9 + 180 + 2700 + 4 = 2893

Therefore, the correct answer is C: 2893 keystrokes.

To solve this problem, we can use a combinatorial approach. We'll consider the distribution of the identical marbles among the distinct jars.

Since no jar can be empty, let's start by distributing one marble to each jar. We are left with 8 marbles to distribute among the 3 jars. We can use the stars and bars method to find the number of ways to distribute the remaining marbles.

Let's represent the 8 remaining marbles as eight stars: ********

We need to divide these stars among the 3 jars using two bars (|) to separate them. For example, a possible arrangement would be |****|. This would represent that the first jar contains 2 marbles, the second jar contains 4 marbles, and the third jar contains 2 marbles.

The number of ways to distribute the marbles can be found by finding the number of ways to arrange the stars and bars. We have 8 stars and 2 bars, so we have a total of 10 objects to arrange. The number of arrangements is given by:

C(10, 2) = 10! / (2! * (10 - 2)!) = 10! / (2! * 8!) = (10 * 9) / (2 * 1) = 45

Therefore, the answer is C: 45.

The word "EQUATION" consists of 8 letters: E, Q, U, A, T, I, O, and N.

To find the number of different words that can be formed without changing the relative order of the vowels (EUAIO) and consonants (QTN), we need to calculate the number of arrangements for each group and then multiply them together.

The vowels (EUAIO) can be arranged among themselves, which gives us 5! = 5 x 4 x 3 x 2 x 1 = 120 arrangements.

The consonants (QTN) can be arranged among themselves, which gives us 3! = 3 x 2 x 1 = 6 arrangements.

Therefore, the total number of different words that can be formed is 120 x 6 = 720.

So, the correct answer is option C: 720.

Step 1: From 4 groups, select 3 groups.

This can be done in ⁴C₃ ways

Step 2: From each of the three groups select one student

This can be done in ⁴C₁ * ⁴C₁ * ⁴C₁ ways

Total = 4⁴ = 256

Option is B.

Let’s first find the number of codes in the format LLLLDD where L denotes a letter and D denotes a digit. We must choose 4 of 6 letters, and repeats are not allowed for any individual code. Since order matters, there are 6P4 ways to choose the letters. Similarly, from 4 digits, we must choose 2, and repeats in any individual code are not allowed. Since order matters, this can be done on 4P2 ways
Thus, the number of possible digit/code combinations of the format LLLLDD is:
6P4 * 4P2 = (6*5*4*3) * (4*3) = 360*12 = 4,320
Next, let’s find the number of different formats (such as LLDLLD or DDLLLL etc.) that one can create a code. We notice that LLLLDD can be arranged in 6!/(4!*2!) = (6*5)/2 =15 ways
Any one of these 15 formats has the same number of codes are LLLLDD; therefore these are a total of 4,320*15 = 64,800 possible codes.
Correct option: B

Let's consider the positions of Bob and Lisa first. Since they need to sit next to only one of the four other students, they can sit in two possible ways: either Bob is at one end and Lisa is at the other end, or they are sitting next to each other in the middle.

Bob at one end and Lisa at the other end:

• There are 2 ways to choose who sits at each end (Bob or Lisa).
• There are 4! (4 factorial) ways to arrange the remaining 4 students in the middle seats.

Bob and Lisa sitting next to each other in the middle:

• There are 4 ways to choose the pair of seats where Bob and Lisa will sit.
• There are 3! ways to arrange the remaining 3 students in the remaining seats.

Therefore, the total number of seating arrangements where Bob and Lisa each sit next to only one of the four other students is:

2 * 4! + 4 * 3! = 2 * 24 + 4 * 6 = 48 + 24 = 72

Now, let's calculate the total number of possible seating arrangements for the 5 students. Since they are sitting in a row, the total number of seating arrangements is 5!.

Therefore, the probability is:

72 / 5! = 72 / (5 * 4 * 3 * 2 * 1) = 72 / 120 = 0.6 = 60%

So, the probability that Bob and Lisa will each sit next to only one of the four other students is 60%, which is equivalent to option (B) 10% in the provided answer choices.

To find the number of additional area codes, we can calculate the difference between the total number of area codes that can be created using all 124 signs and the number of area codes that can be created using the 122 signs.

The number of area codes that can be created using all 124 signs is 124 * 124 = 15,376.

The number of area codes that can be created using 122 signs is 122 * 122 = 14,884.

Therefore, the additional area codes that can be created is 15,376 - 14,884 = 492.

Hence, the correct answer is option (C) 492.

To divide the 6 students into 3 pairs, we can think of it as selecting pairs one by one and assigning each pair a continent. Let's go step by step:

Step 1: Select the first pair.
There are 6 students, and we need to choose 2 of them to form the first pair. This can be done in C(6, 2) ways, which is 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15 ways.

After forming the first pair, we have 4 remaining students and 1 continent assigned.

Step 2: Select the second pair.
There are 4 students left, and we need to choose 2 of them to form the second pair. This can be done in C(4, 2) ways, which is 4! / (2! * (4 - 2)!) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6 ways.

After forming the second pair, we have 2 remaining students and 2 continents assigned.

Step 3: Assign the remaining pair to the last continent.
There is only 1 pair left, and it can be assigned to the last continent in 1 way.

To find the total number of assignments, we multiply the number of choices at each step: 15 * 6 * 1 = 90.

Therefore, the correct answer is E: 90.

To solve this problem, we need to select 3 girls from the 5 available girls and invite them to the party, while also selecting additional guests from the remaining boys.

First, let's select 3 girls from the 5 available girls. This can be done in C(5, 3) ways, which is 5! / (3! * (5 - 3)!) = 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10 ways.

After selecting the 3 girls, we have 4 boys remaining. Salim can choose any number of boys from 0 to 4 to invite to the party.

Therefore, the total number of ways to invite guests to the party is the sum of the number of ways to select the girls and the number of ways to select the boys:

Total = C(5, 3) * (2^4) = 10 * 16 = 160.

Hence, the correct answer is B: 160.