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In a carom tournament in school all the kids played against each other exactly once. The number of games involving a girl and a boy is 150 and the number of games involving two girls is 60 less than the number of games involving two boys, find the total strength of the school?
Let the number of boys in the class= B
number of girls in the class = G.
"The number of games involving a girl and a boy is 150" ==> Bc1*Gc1 = 150 ==> B*G = 150 --- (1)
"the number of games involving two girls is 60 less than the number of games involving two boys" ==> Gc2 +60 =Bc2 ==> G(G-1) + 120 = B(B-1) --- (2)
By solving equations (1), (2) we can deduce the values of B, G.
B=15, G=10.
So the total strength of the class = 10+15 = 25.
ABC Paints Ltd. is planning to create different combination of dyes. The research team has decided they will be using five different green dyes, three different red dyes and four different blue dyes. How many combinations of dyes can be created by ABC Paints Ltd., by including at least one blue and one green dye?
Number of green dyes = 5
Number of blue dyes = 4
Number of red dyes = 3
The no of ways of choosing of at least one green and one blue is = $$\left(2^5-1\right)\left(2^4-1\right)2^3$$
=31*15*8
= 3720
B is the correct answer.
Sonal and Meenal appear in an interview for same post having two vacancies. If $$\frac{1}{7}$$ is Sonal's probability of selection and $$\frac{1}{5}$$ is Meenal's probability of selection then what is the probability that only one of them is selected?
Probability of Sonal's selection = $$\frac{1}{7}$$
Probability of Meenal's selection = $$\frac{1}{5}$$
The probability that only one of them is selected = $$\frac{1}{7}$$*$$\frac{4}{5}$$ + $$\frac{6}{7}$$*$$\frac{1}{5}$$
= $$\frac{10}{35}$$
= $$\frac{2}{7}$$
B is the correct answer.
Big Bang Theory cast wishes to find out the number of ways in which the word ASTRONAUT can be scrambled. They find that the number of ways in which it can be put in an unscrambling puzzle is?
Number of letters in ASTRONAUT = 9
Here A and T are repeated twice.
So the number of ways of arrangement = $$\ \frac{\ 9!}{2!\times\ 2!}$$
= 90720
D is the correct answer.
How many different words can be formed with the word CUSTOM with a condition that the word should begin with M?
Assume that all words have 6 distinct letters.
Number of different words that can be formed using the letters of the word CUSTOM and starting with M are
M,_, _, _, _, _.
These 5 blanks have to be filled with the remaining 5 letters in 5*4*3*2*1 = 120 ways.
C is the correct answer.
The number of ways that 5 Marathi, 3 English and 3 Tamil books be arranged if the books of each language are to be kept together is
Marathi, English and Tamil books can be arranged in 3! ways.
All the Marathi books can be arranged among themselves in 5! ways.
Similarly, the English and Tamil books can be arranged among themselves in 3! ways.
The total number of ways in which the books can be arranged = 3!*5!*3!*3!
=25920
A is the correct answer.
How many words each of two vowels and three consonants can be formed from the letters of the word "UNIVERSAL" ?
The number of ways of selecting 2 vowels from U,I,A,E = $$4_{c_2}$$ = 6
The number of ways of selecting 3 consonants from N,V,R,S,L = $$5_{c_3}$$ = 10
After selecting you can arrange them is 5! ways.
Total number of words = 6*10*120 = 7200
Sonali can solve 70% of the problems in a competitive exam and Nirali can solve only 60% in the same exam. What is the probability that at least one of them will solve a problem, provided selection of questions is done randomly from the same exam ?
P(atleast 1 solved) = 1 - P(no one solved)
= 1-(1-0.7)x(1-0.6)
= 1 - (0.3)x(0.4)
= 1 - 0.12
= 0.88
A person has a bag which contains 9 bulbs out of which 2 are fused and cannot be used to lighten the room. Two bulbs are selected at random. What is the probability that all the two bulbs chosen can be used to lighten the room ?
Probability = Number of favourable cases/ total number of cases
favourable cases= $$7_{c_2}$$
total cases = $$9_{c_2}$$
Probability = $$\frac{\left(7c_2\right)}{\left(9_{c_c}\right)}$$ =7/12
There are nine humans in a ship, each human has nine cages and each cage has nine huge lions and each lion has nine cubs. How many legs are there in the ship ? (Human have two legs, lions have four legs, cubs have four legs.)
Total number of humans = 9 => total legs = 9*2=18
Total number of lions = 9*9*9 => total legs= 9*9*9*4=2916
Total number of cubs = 9*9*9*9=> total legs =9*9*9*9*4=26244
So total legs= 29178
How many different letter arrangements can be made from the letter of the word EXTRA in such a way that the vowels are always together?
Considering two vowels as one unit ,total number of units = 4
Number of way of arranging these 4 units = 4! = 24 ways
2 vowels can be internally arranged in 2! ways 2 ways
Therefore total number of ways = 24*2 =48 ways.
In a given race the odds in favour of three horses A, B, C are 1:3; 1:4; 1:5 respectively. Assuming that dead heat is impossible the probability that one of them wins is
Odds in favour of A = 1:3, thus P(A wins) = $$\frac{1}{1+3}=\frac{1}{4}$$
Odds in favour of B= 1:4, thus P(B wins) = $$\frac{1}{1+4}=\frac{1}{5}$$
Odds in favour of C = 1:5, thus P(C wins) = $$\frac{1}{1+5}=\frac{1}{6}$$
Since deadheat is not possible only 1 will win.
P(A or B or C wins) = $$\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}$$
A bag contains 5 white and 3 black balls, and 4 are successively drawn out and not replaced. What’s the chance of getting different colours alternatively?
Probability of alternate balls when first ball is black = $$\frac{5}{8}\times\ \frac{3}{7}\times\ \frac{4}{6}\times\ \frac{2}{5}=\frac{1}{14}$$
Probability of alternate balls when first ball is white = $$\frac{3}{8}\times\ \frac{5}{7}\times\ \frac{2}{6}\times\ \frac{4}{5}=\frac{1}{14}$$
Overall Probability = $$\frac{1}{14}+\ \frac{1}{14}=\frac{1}{7}$$
A bag contains 100 tickets numbered 1, 2, 3, .... 100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?
We will first look at the numbers that have 2 in the unit's digit.
So, the numbers between 1 to 100 having 2 in unit's place are 2, 12, 22, 32... 92. So, there are 10 such numbers.
For numbers having 2 at ten's place, we have numbers from 20 to 29. In total we have 10 such numbers.
But, 22 is common to both the groups, and hence,
Total numbers with between 1 and 100 with digit 2 on it is 10+10-1=19.
Total sample space= 100.
So, required probability= $$\ \frac{\ 19}{100}$$
There are five boys and three girls who are sitting together to discuss a management problem at a round table. In how many ways can they sit around the table so that no two girls are together?
The five boys can be arranged on a round table in (5-1)! ways i.e 24
Now there will be five spaces created between two boys.
So three girls can be seated in $$^5C_3$$ ways*3! = 10*6=60
Total number of ways in which five boys and three girls can be seated = 24*60 = 1440
D is the correct answer.
The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a meeting consisting of 8 ladies and 7 gentlemen, if Mrs. X refuses to serve in a committee if Mr. Y is its member, is
Here we have two cases:
Case 1: If Y is a member of the committee
We have to select 3 other gentlemen and 3 ladies from 8 ladies and 5 gentlemen (Since If Y is a member of the committee, X will not be a member of the committee)
=$$^7C_3\times\ ^6C_3$$ =35*20=700.
Case 2: If Y is not a member of the committee
We have to select 4 gentlemen and 3 ladies from 8 ladies and 6 gentlemen (Since Y is not a member of the committee)
=$$^8C_3\times\ ^6C_4$$ = 56*15=840
Total cases = 700+840 = 1540
C is the correct answer.
A family consists of a grandfather, 5 sons and daughters and 8 grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each end and the grandfather refuses to have a grandchild on either side of him. The number of ways in which the family can be made to sit is
Total no. of seats = 1 grandfather+ 5 sons and daughters + 8 grandchildren
= 14.
The grandchildren can occupy the 4 seats on either side of the table in $$^8P_4$$*4! ways
The grandfather can occupy a seat in (5-1)= 4 ways (4 gaps between 5 sons and daughter).
The remaining seats can be occupied in 5!= 120 ways (5 seats for sons and daughter).
Total number of required ways = 8!*4*120
=8!*480
D is the correct answer.
At a college football game, $$\frac{4}{5}$$ of the seats in the lower deck of the stadium were sold. If $$\frac{1}{4}$$ of all the seating in the stadium is located in the lower deck, and if $$\frac{2}{3}$$ of all the seats in the stadium were sold, then what fraction of the unsold seats in the stadium was in the lower deck?
Let the number of seats in the stadium = x
Number of seats in the lower deck = a and the number of seats in the upper deck be b
x=a+b
a=$$\ \frac{\ x}{4}$$ and b = $$\ \frac{\ 3x}{4}$$
Now in the lower deck $$\ \frac{\ 4a}{5}$$ seats were sold and $$\ \frac{\ a}{5}$$ seats were unsold.
Number of total seats sold in the stadium = $$\ \frac{\ 2x}{3}$$
Number of unsold seats in the lower deck = $$\ \frac{\ a}{5}$$ = $$\ \frac{\ x}{20}$$
Number of unsold seats in the stadium = $$\ \frac{\ x}{3}$$
Required fraction = $$\ \frac{\ \ \frac{\ x}{20}}{\ \frac{\ x}{3}}$$
= $$\ \frac{\ 3}{\ \ 20}$$
A is the correct answer.
The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, is:
Total number of days in a leap year = 366
It will contain 52 weeks and 2days
These two days can be (Sunday, Monday); (Monday, Tuesday); (Tuesday, Wednesday); (Wednesday, Thursday); (Thursday, Friday); (Friday, Saturday); (Saturday, Sunday)
For 53 Sundays, probability = $$\ \ \frac{2}{7}$$
Similarly for 53 Mondays, probability =$$\ \ \frac{2}{7}$$
This includes one way where Sunday and Monday occur simultaneously (i.e) Sunday, Monday
Probability for this =$$\ \ \frac{1}{7}$$
Hence required probability = $$\ \ \frac{2}{7}$$ +$$\ \ \frac{2}{7}$$-$$\ \ \frac{1}{7}$$
=$$\ \ \frac{3}{7}$$
C is the correct answer.
A bag contains 5 white and 3 black balls; another bag contains 4 white and 5 black balls. From any one of these bags a single draw of two balls is made. Find the probability that one of them would be white and another black ball.
Let P(B1) and P(B2) denote the probabilities of the events of drawing balls from the bag I or bag II.
Since both the bags are equally likely to be selected, P(B1) = P(B2) = $$\ \frac{\ 1}{2}$$
Let E denote the event of drawing two balls of different colours from a bag.
The required probability
= P(B1)P($$\ \frac{\ E}{B_1}$$)+ P(B2)P($$\ \frac{\ E}{B_2}$$)
P($$\ \frac{\ E}{B_1}$$) denote the probability of drawing the two balls from bag I and as that from bag II.
= $$\ \frac{\ 1}{2}\left[\ \frac{\ ^5C_1\ \times\ \ ^3C_1}{^8C_2}\ +\ \ \frac{\ ^5C_1\ \times\ \ ^4C_1}{^9C_2}\right]$$
= $$\ \frac{\ 1}{2}\left[\ \frac{\ 15}{28}+\ \frac{\ 5}{9}\right]$$
= $$\ \frac{\ 275}{504}$$
A is the correct answer.
If $$^nC_x = 56$$ and $$^nP_x = 336$$, then Find n and x?
$$^nP_r$$ = $$^nC_r$$*r!
$$^nP_x = 336$$
$$^nC_x *x!$$ = 336
x!= $$\ \frac{336}{56}$$
x!=6
x=3
$$^nC_x = 56$$
$$\ \ \frac{\ n!}{\left(n-x\right)!\times\ x!}\ =\ 56$$
$$\ \ \frac{\ n!}{\left(n-3\right)!\times\ 3!}\ =\ 56$$
n=8
$$\therefore$$ n= 8, x=3
C is the correct answer.
A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15. The chance that the first roll was a four is
There are three possibilities for sum of uppermost faces of dice is equal to 15. They are
5, 5, 5 - only 1 arrangement
4, 5, 6 - 3! = 6 arrangements
3, 6, 6 - $$\frac{3!}{2}$$ = 3 arrangements
Out of these 10 arrangements, there are 2 cases in which 4 occurs in first roll ( 4,5,6 or 4,6,5)
Therefore, probability = $$\frac{2}{10}$$ = $$\frac{1}{5}$$
Answer is option B.
In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
ABACUS
Vowels = AAU
Remaining BCS
Now AAU will be considered 1 and B,C,S separate
so these can be arranged in 4! ways
Vowels AAU can be arranged in 3!/2! ways
so total arrangements = $$\frac{4!\times\ 3!}{2!}$$
A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:
Let us observe the digits given to us, we have 0,1,2,3,4 and 5, to make a five digit number we only need 5 digits out of the given 6 digits,
Case 1: Using digits 0,1,2,4 and 5.
The number of ways in which we can arrange these 5 digits are ⇒4×4×3×2×1
⇒96
Case 2: Using the digits 1,2,3,4 and 5
The number of ways in which we can arrange these 5 digits are ⇒5×4×3×2×1
⇒120
Therefore, the total number of cases =96+120=216
A five digit number is formed by using the digits 1, 2, 3, 4 and 5 without repetitions. What is the probability that the number is divisible by 4?
For a number to be divisible by 4, last 2 digits must be divisible by 4
Last 2 digits can be 12, 24, 32, 52
_, _, _, _, _
Last 2 digits have 4 possibilities and remaining digits can be arranged in 3! ways
Number of 5 digit numbers divisible by 4 are 3! $$\times\ $$ 4 = 24
Total number of 5 digit numbers that can be formed = 5!
Therefore, probability = $$\frac{24}{5!}\ $$ = $$\frac{1}{5}$$
Answer is option A.
There are 10 stations on a railway line. The number of different journey tickets that are required by the authorities is
The number of different journey tickets that are required by the authorities = The number of selecting 2 stations × 2 (For every pair of stations, say A and B, there are two tickets required A to B and B to A)
= $$^{10}C_2\times\ 2$$ = 90
There are 10 stations on a railway line. The number of different journey tickets that are required by the authorities is
The number of different journey tickets that are required by the authorities = The number of selecting 2 stations $$\times\ $$ 2 (For every pair stations say A and B, there are two tickets required A to B and B to A)
= $$\\^{10}C_2$$ * 2 = 45*2=90
A and B throw one dice for a stake of Rs.11, which is to be won by the player who first throws a six.
The game ends when stake is won by A or B. If A has the first throw, what are their respective expectations?
The probability of throwing a 6 = $$\ \frac{\ \ 1}{\ 6}$$
The probability of not throwing a 6 = $$\ \frac{\ \ 5}{\ 6}$$
The probability that A wins = $$\ \frac{\ \ 1}{\ 6}+\ \frac{\ 5}{\ 6}\times\ \ \frac{\ 5}{6}\times\ \ \frac{\ 1}{6}+\ \frac{\ 5}{6}\ \times\ \ \frac{\ 5}{6}\times\ \ \frac{\ 5}{6}\times\ \ \frac{\ 5}{6}\times\ \frac{\ 1}{6}\ +\ ...............$$
= $$\ \frac{\ 1}{6}\left(1+\left(\ \frac{\ 5}{6}\right)^2+\left(\ \frac{\ 5}{6}\right)^4.........\right)$$
= $$\ \ \frac{\ 1}{6}\left(\ \dfrac{\ 1}{\ 1-\ \frac{\ 25}{36}}\right)$$ = $$\ \frac{\ 6}{11}$$
The expected return of A = 11*$$\ \frac{\ 6}{11}$$ = 6
Hence the expected return of B = 11-6=5
Four stacks containing an equal number of chips are to be made from 11 orange, 9 white, 13 black and 7 yellow chips. If all of these chips are used and each stack contains at least one chip of each colour, what is the maximum number of white chips in any one stack?
Total number of chips = 11+9+13+7=40
Total number of chips in each stack = 40/4 = 10
Total number of white chips = 9,
The maximum number of white chips in a stack will be obtained when 1 white chip given to each of 3 stacks and 6 white chips to 1 stack.
Hence, 6 is the answer.
A number lock consists of 3 rings each marked with 10 different numbers. In how many cases the lock cannot be opened?
The total number of combinations = 10*10*10 = 1000 (There are 10-digits 0 to 9 on each ring)
Since there is only 1 right combination, the number of cases in which the lock cannot be opened = 1000-1=999
Thirty days are in September, April, June and November. Some months are of thirty one days. A month is chosen at random.
Then its probability of having exactly three days less than maximum of 31 is
Number of days in February = 28
Number of months having three days less than 31 i.e, 28 days = 1
$$\therefore\ $$Required probability = $$\frac{1}{12}$$
Hence, the correct answer is Option D
A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are 100 Year- Ill, 150 Year-II, and 200 Year-I students who applied. Each Year-III’s name is placed in the lottery 3 times; each Year-II’s name, 2 times; and each Year-I’s name, 1 time. What is the probability that a Year-III’s name will be chosen?
By the given information,
Total number of Year III student's name= 3*100= 300
Total number of Year II student's name= 2*150= 300
Total number of Year I student's name= 1*200= 200.
Number of favourable outcomes= year III student's name= 300
Total Sample space= total number of names= 300+300+200= 800.
Therefore, required probability= $$\ \frac{\ 300}{800}=\ \ \frac{\ 3}{8}$$- Option D
How many arrangements can be formed out of the letters of the word EXAMINATION so that vowels always occupy odd places?
Number of vowels in word EXAMINATION =6
Number of consonants in word EXAMINATION =5
Number of ways in which 6 vowels can be arranged in 6 odd places=(6!)/2!*2! (1)
Number of ways in which 5 consonants can be arranged in 5 even places = 5!/2! (2)
Now total arrangements = (1)*(2)
=10,800
In a school drill, a number of children are asked to stand in a circle. They are evenly spaced and the 6th child is diametrically opposite the 16th child. How many children are made to stand in the circle?
The 6th and the 16th child divide the circle into two halves as it is given that the 6th child is diametrically opposite the 16th child.
When we consider one of the two halves, it will contain students numbered from 7 to 15, i.e a total of 9 children.
Similarly, the other side must also have 9 children.
Now, including the two pairs of 9 children at each half along with child numbered 6 and 16, we have a total of 9+9+2= 20 children.
128 players start in the men's single at a tennis tournament, where this number
reduces to half on every succeeding round. How many matches are played totally in the event?
In Round 1 : 64 matches are played among 128 members
In Round 2: 32 matches are played among 64 members
In Round 3 : 16 matches are played among 32 members
and so on till 1 match is played among final 2 Players.
Total matches = 1+2+4+8+16+32+64 = 127
If Swamy has two children and he truthfully answers yes to the question "Is at least one of your children a girl?" what is the probability that both his children are girls?
Probability = $$\frac{\left(Number\ of\ favourable\ cases\right)}{Total\ number\ of\ cases}$$
Here the possible outcomes are GG or GB
So probability of two girls = 1/2
There are 6 tickets to the theater, four of which are for seats in the front row. 3 tickets are selected at random. What is the probability that two of them are for the front row?
Out of all the 6 tickets, 4 are front row tickets and the rest 2 are not.
Out of 4 front tickets 2 can be selected in $$^{4\ }C\ _2$$
The remaining 1 ticket has to be selected in $$^{2\ }C\ _1$$
Hence total ways of selecting 3 tickets such that 2 of them are front row are $$^{4\ }C\ _2$$ * $$^{2\ }C\ _1$$
3 out of 6 tickets can be selected in $$^{6\ }C\ _3$$ ways.
Required probability = $$^{4\ }C\ _2$$ * $$^{2\ }C\ _1$$ / $$^{6\ }C\ _3$$
Which is equal to 0.6
You are given 50 white marbles, 50 black marbles and two jars. You need to put 100 marbles in any of these two jars. The jars will then be shaken and you will be asked to pick one marbles from either jar. How would you distribute the marbles in two are to maximize the possibility of picking a white marble blind folded?
The probability of picking the white is 0.5*(Probability of white in Jar1) + 0.5*(Probability of white in Jar2)
To maximize the probability we can put 1 white marble in jar 1. Hence the probability of white in jar 1 = 1
The remaining 49 white and 50 black can be put in Jar 2. Thus, the Probability of white in Jar 2 = 49/99.
All other options give less probability
SNAP Probability and Combinatorics is an important topic in the Quantitative, Data Interpretation & Data Sufficiency section of the SNAP 2026 Exam. Questions from this topic test a candidate's understanding of probability, permutations, combinations, counting principles, and event-based reasoning. These concepts help candidates solve problems involving arrangements, selections, and the likelihood of outcomes. Practicing topic-wise questions and solving problems from SNAP Previous Papers is an effective way to strengthen concepts and improve accuracy. Additionally, attempting SNAP Mock Tests will help candidates understand the latest exam pattern and the types of questions that can be expected in the actual exam. To gain a better understanding of all the topics covered in the exam, candidates should also refer to the SNAP Syllabus.
Download SNAP Previous Papers
Yes, Probability and Combinatorics are important topics in SNAP Quantitative Aptitude. These topics test logical reasoning, analytical ability, and mathematical problem-solving skills, making them valuable for scoring well in the exam.
SNAP does not specify a fixed number of questions from Probability and Combinatorics. However, candidates can expect direct or concept-based questions from probability, permutations, combinations, and counting principles.
SNAP may include questions on probability of events, conditional probability, permutations, combinations, arrangements, selections, and counting techniques. Questions often focus on applying concepts efficiently rather than performing lengthy calculations.
Start by understanding the fundamental concepts of probability, permutations, and combinations. Practice a variety of questions, solve problems from SNAP Previous Papers, and take SNAP Mock Tests regularly to improve speed and accuracy.
The difficulty level is generally easy to moderate. While some questions may require careful logical analysis, most can be solved quickly with strong conceptual understanding and regular practice.
Cracku's SNAP Probability and Combinatorics Questions are designed according to the latest SNAP exam pattern and difficulty level. They provide topic-wise practice questions, detailed solutions, shortcut methods, and performance analysis to help aspirants strengthen their concepts and improve their scores.
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