GMAT Sets and Counting Method Questions With Solutions PDF

The number of ways of picking a vowel is 5 and the number of ways of picking a consonant is 4.

The remaining 7 letters can be arranged in 7! ways.

Hence, the total number of arrangements is 5*7!*4 = 100800 ways.

Total number of ways = $$^5 c_5 * ^6 c_1 + ^4 c_4 * ^5 c_2 = 6+10 = 16$$ ways

Total number of ways
$$=^4 c_3 * ^6 c_3 +^4 c_3*^5 c_1*^6 c_2 + ^4 c_4* ^6 c_2 +^4 c_4 *^5 c_1*^6 c_1 $$
$$=4 \times 20+4 \times 5 \times 15+15+5 \times 6$$
$$=80+300+15+30=425$$ ways.
Choice (c)

In ‘LOGARITHMS’ Total number of letters = 10 No’s
Out of 10 letters, 3 letter can be formed = $$^{10}P_3$$ = (10×9×8×7!)/7! =720

The word “School” has 6 letters but “O” repeats twice. So, the number of ways to arrange it are 6!/2! = 360 ways.

The word “Question” has 8 letters. So, the number of ways to arrange it are 8! = 40,320 ways.

The word “Security” has 8 letters. So, the number of ways to arrange it are 8! = 40,320 ways.

The total number of letters is 6 and no letter is repeated. So, the total number of arrangements is 6! = 720 ways.

Total no. of ways arranging all 7 letters of the word = 7!
Total no. of ways arranging letters with combining both vowels together = 6!*2
Ways of arranging letters when no two vowels come together = 7! - 6!*2 = 3600

No. of ways arranging letters will be = 6!/2 (As 2 “T” are there)

Total letters of the word = 10 with 2 A’s
Hence, no. of ways of arranging them = 10!/2

As there are 8 letters with 2 U’s hence ways of arranging will be 8!/2

(Total no. of ways arranging them) - (Ways of arranging them with 2 vowels)
6! - 5!(2) = 5!(6-2) = 4(5!)

As in the given word “COURAGE”, number of different letters are 7.
So, ways of arranging 7 different letters will be = 7!

In “PRESENT” among 7 letters two E’s are there.
Hence, total no. of ways to arrange it will be = 7!/2! = 2520

As we know that there are 2N’s in the word NATION with total of 6 letters
Hence, total number of ways of arranging them will be = $$ \frac{6!}{2!} = 360 $$

Probability that both the balls are red = $$\frac{^4C_2}{^{12}C_2} = \frac{1}{11}$$

We have to select 2 trainees out of 3 trainees and 3 resarch associates out of 6 resarch associates.

Required number of ways = $$^3C_2*^6C_3 = 60$$ ways

There are 6 letters in the word of which one of them appears twice. So, the number of arrangements would be $$ \frac{6!}{2!}$$ = 360

There are 7 letters in the word of which one of them appears twice. So, the number of arrangements would be $$ \frac{7!}{2!}$$ = 2,520

The total number of letters in the word 'DOORS' is 5.
In this word, the alphabet 'O' is repeated twice.
Hence, the total number of ways of arranging the letters of the word 'DOORS' equals 5!/2 = 5*4*3*2*1/2 = 60

The total number of letters in the word 'SUPERB' is 6.
As there are no repetitions, the total number of ways of arranging the letters is 6! = 6*5*4*3*2*1 = 720

The vowels in the word are: A, I, E, I.
The consonants are: R, T, H, M, T, C

Consider the vowels as a single unit. There are a total of 7 units of which 2 Ts are similar. So, they can be arranged in 7!/2! ways. The vowels themselves can be arranged in 4!/2! ways = 12 ways
So, the total number of arrangements = 7!/2! * 4!/2! ways = $$(7!*4!)/(2!^2)$$

Number of people to be selected in the committee = 10
Number of people available for selection = 20
So, number of ways in which the people can be selected = $$^{20}C_{10}$$

The first 5 whole numbers are 0, 1, 2, 3, 4
Since we have to form a 5 digit number, the first digit cannot be 0.
Number of ways = 4*4*3*2*1= 96

Let us group 2 E's and arrange this group along with 7 other letters.
Number of ways = 8!
Number of waus of arranging 2 E's in a group = 1
Net number of ways = 8!

Lets say there are n people. Now each person shakes hands with
n-1 people, total number of handshakes = n*(n-1). However, each handshakes involves 2 people so when we are doing n*(n-1), we are counting each handshake twice. So total number of handshakes are given by $$ \frac{ n*(n-1)}{2}$$
Given $$\frac{ n*(n-1)}{2} = 36 $$=> $$n^{2}-n =72$$ => $$n^{2}-n-72 = 0$$
So on solving this equation we get n =9,-8 . Since n is the number of people, it can’t be negative. So, n =9. Thus, option A is correct.

The number of letters in HYDERABAD = 9
There are 2 A’s and 2 D’s
There are 3 vowels. If all these are together then they can be considered as 1 unit.
So they can be arranged in 7!/2! = 2520
Now 3 vowels can have internal arrangement among them which can happen in
3!/2! = 3 ways
So required number of ways = 2520*3 = 7560

The alphabetical order of the letters is AALMR.

Number of words beginning with AA = 3! = 6
Next would come ALAMR and then ALARM.
Hence, rank would be 8.

The number of letters in INFINITY = 3 Is + 2Ns + 1 F + 1T + 1Y
If we treat all the three I's as one element, we get 6 elements of which there are 2 Ns.
So, the number of ways of rearranging them = $$\frac{6!}{2!}$$ = 360 ways

Since, the number formed should be divisible 12, the number should be divisible by 3 and 4.
Since, the number is divisible by 3, the sum of digits should be a multiple of 3
The only possible combinations of 4 which is a multiple of 3 is (1,2,4,5)
Since, the number is divisible by 4, the last two digits can only be 12, 24 and 52.
If the last two digits is 12, then there are 2! ways of arranging the first two digits.
Thus, since there are three different combinations of last two digits, the required number of ways = 2! X 3 = 6

The word COMMUNICATION has 13 letters in it with each of M, N, C, O andI repeating twice.
Therefore the number of arrangements possible = 13!/(2!*2!*2!*2!*2!) = $$13!/2^5$$

5 batsmen can be selected in 6C5 = 6 ways
1 wicket keepers can be selected in 2C1 = 2 ways
2 all rounders can be selected in 3C2 = 3 ways
3 bowlers can be selected in 5C3 = 10 ways
Hence the required number of ways are 6*2*3*10 = 360

Total number of ways of arranging 6 people around the circular table will be (n-1)! [here n=6]
=> No. of ways = 5!
Let’s consider the two females as one object.
No. of arrangements in which the females are together = 4! X 2 [Multiplied by 2 as 2 females can be arranged in 2 ways]
So, the number of arrangements where 2 females don’t sit next to each other = 5! - 4! X 2 = 4! X 3 = 72 ways.

Let the four digit even number be in the form abcd.
a, b, c can take 5 values each
d can take only one value
Therefore, the number of four digit even numbers possible = 5*5*5*1 =125.

Here, 15 can be obtained in following cases {5,5,5},{6,5,4},{6,6,3}
Number of possibilities of getting a {5,5,5} outcome = 1
Number of possibilities of getting a {6,5,4} outcome = 3! = 6 (these 3 outcomes can be arranged on A,B,C)
Number of possibilities of getting a {6,6,3} outcome = $$^3c_1 $$ {getting a 3 on one of the three dices}
Hence, there are (1+6+3)=10 cases.

3 bowlers out of 5 can be selected in 5C3 ways = 10 ways
1 keeper out of 3 can be selected in 3C1 ways = 3 ways
4 batsman out of 7 can be selected in 7C4 ways = 35 ways
3 allrounder out of 4 can be selected in 4C3 ways = 4 ways
So total WAYS = 10*3*35*4 = 4200 Ways
So E is the right choice.

The four numbers can be arranged in 4! ways=24 ways. Hence there are 24 possible pin numbers and the thief would need a maximum of 24 tries.

Expected value is the probability weighted average of all expected outputs.
Let's list out the expected outputs and their probabilites
1) All questions wrong:. score = -3 , probability = $$0.7^3 $$
2) 1 Question right: score = +3 -2 = 1. Probability = $$ ^3 C _1 * 0.7^2*0.3$$
3) 2 Questions right: score = +6 -1 = 5. Probability = $$ ^3 C _2 * 0.7*0.3^2$$
4) All Question right: score = +9. Probability = $$ ^3 C _1 * 0.3^3$$

The expected value is given by $$-3 * 0.7^3 + 1 * ^3 C _1 * 0.7^2*0.3+ 5*^3 C _2 * 0.7*0.3^2 + 9 *^3 C _3*0.3^3 = 0.6$$

The number of such numbers that can be formed are 5 X 5 X 5=125.

This means that we have 125 unit digits, 125 tens digits, and 125 hundred digits. At each of these places, every number is equally likely to appear. This means that at each of unit, tens, and hundred place, each number will appear for 125/5=25 times.

Now, the total sum of digits for all such numbers will be- 25X(1+3+5+7+9) (for hundred) + 25X(1+3+5+7+9) (for tens)+ 25X(1+3+5+7+9) (for unit place)= 25 X 3 X (1+3+5+7+9) = 25 X 3 X 25. Now, the expected value will be the average of the sum, i.e. 25 X 3 X 25/125= 15.