Let's take the number of paths between Q and R to be b and the number of paths between R and S to be a<img 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">We are given the paths from P to S through R (which would be 4a), the paths from P to S&nbsp;through Q (which would be 12) and the paths from P to Q to R to S, which would be 3ab) is equal to 62Giving the relation 4a+12+3ab = 62Or 4a+3ab = 50The paths from Q to R directly (which would be b), through P( which would be 12) and through S (which would be 4a) are 27Giving the relation b+12+4a = 27Or 4a+b = 15Subtracting this equation from the first one we got, we get 3ab-b=35, or b(3a-1)=35b can be 1, 3, 5 or 7Substituting these values in the second equation, we see that it can not be 1 or 5, leaving only 3 or 7 as the possible values.&nbsp;Substituting b as 3 in the first equation would give 13a=50, which is not true.&nbsp;Substituting b as 7 in the first equation would give 25a= 50, which would give a=2We are asked the number of paths from Q to R, which is b=7Therefore, 7 is the correct answer.&nbsp;

Let the number of balloons each child received be 2a, 2b, 2c and 2d2a + 2b + 2c + 2d = 20a + b + c + d = 10Each of them should get more than zero balloons.Therefore, total number of ways =&nbsp;$ (n-1)C_{r-1} = (10-1)C_{4-1} = 9C_3 = 84 $

Case 1: 4-digit numbersGiven digits - 0, 1, 2, 3, 4, 5_, _, _, _As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 waysTotal number of possible 4-digit numbers = 4*60 = 240Case 2: 5-digit numbers_, _, _, _, _First digit cannot be zero.Therefore, total number of cases = 5*5*4*3*2 = 600Case 3: 6-digit numbers_, _, _, _, _, _First digit cannot be zero.Therefore, total number of cases = 5*5*4*3*2*1 = 600Total number of integers possible = 600 + 600 + 240 = 1440The answer is option A.

The number of 4-digit numbers possible using 1,1,2, and 4 is $\frac{4!}{2!} = 12$Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.Sum = $6(1) + 3(2) + 3(4) = 24$Similarly, in tens digit, hundreds digit and thousands digit as well.Therefore, sum = $24 + 24(10) + 24(100) + 24(1000) = 24(1111)$Mean = $\frac{24(1111)}{12} = 2222$

The possible arrangements are of the form&nbsp;35 _ Can be chosen in 6 ways.35 _ _&nbsp;We can choose 2 out of the remaining 6 in&nbsp;6C2=15 ways.We remove 1 case where 7 and 8 are together to get 14 ways.35 _ _ _We can choose 3 out of the remaining 6 in&nbsp;6C3=20 ways. We remove 4 cases where 7 and 8 are together to get 16 ways.35 _ _ _ _We can choose 4 out of the remaining 6 in&nbsp;6C4=15ways. We remove 6 case where 7 and 8 are together to get 9 ways.35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.Thus, total number of cases = 6+14+16+9+2 = 47.Alternatively,The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.Considering the cases, only 7 is selected.We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.Hence we must have ( 3, 5, 7) for the remaining 4 numbers we haveEach of the numbers can either be selected or not selected and we have 4 numbers :Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.SImilarly, including only 8, we have 16 more possibilities.Cases including neither 7 nor 8.We must have 3 and 5 in the group but there must be no 7 and 8 in the group.Hence we have 3 5 _ _ _ _.For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.= 16 possibilitiesBut we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.Hence 16 - 1 = 15 cases.Total = 16+15+16 = 47 possibilities

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.Then (100c+10b+a)-(100a+10b+c)=19899c-99a=198c-a = 2.Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.Thus, there can be 7 cases.B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be&nbsp;7×&nbsp;10=70

Event 1: Distribution of balloonsSince each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.So we are left with $15 - 4 \times 3 = 15 - 12 = 3$ balloons and 3 children.Now we need to distribute 3 identical balloons to 3 children.This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$Event 2: Distribution of pencilsSince each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with $6 - 3 = 3$ pencils.We now need to distribute 3 identical pencils to 3 children.This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$Event 3: Distribution of erasersWe need to distribute 3 identical erasers to 3 children.This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$Applying the product rule, we get the total number of ways = $10 \times 10 \times 10 = 1000$.

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.The different possibilities include :Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :$\frac{4!}{3!​}$&nbsp;= $4$.Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :$\frac{4!}{2!(2!)​}$&nbsp;= $6$.Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :$\frac{4!}{3!​}$&nbsp;= $4$.Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :$\frac{4!}{2!​}$&nbsp;= $12$.Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :$\frac{4!}{2!​}$&nbsp;= $12$.Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :$\frac{4!}{2!​}$&nbsp;= $12$.A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities. Alternatively,We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once&nbsp;So the possible cases :<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/4.image_cXlcGFJ.png">Now we get&nbsp;$\frac{4!}{2!​}$&nbsp;×&nbsp;3 = 36 ( When one digit is used twice and the remaining two once )$\frac{4!}{3!​}$&nbsp;×&nbsp;2&nbsp;= 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )$\frac{4!}{2!​ × 2!}$&nbsp;=&nbsp;6( When 2 and 3 is used 2 times each )So total numbers = 36+8+6 = 50

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)To calculate&nbsp;the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y directionHence&nbsp;the number of paths from (1,1) to (4,6) =&nbsp;${}^{(3+5)}C_3​$&nbsp;= 56To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y directionHence the number of paths from (4,6) to (8,10) =&nbsp;${}^{(4+4)}C_4$​&nbsp;= 70The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number. There are 9 numbers (1,2,3,4,5,6,7,8,9) in total. 2 digit numbers can be formed in $9C_2$ ways. 3 digit numbers can be formed in $9C_3$ ways. .................................................. ..9 digit number can be formed in 9C9 ways.&nbsp; We know that $nC_0+nC_1+nC_2+.........+nC_n =2^n$ =&gt; $9C_0 + 9C_1+9C_2+....9C_9 = 2^9$ $9C_0 + 9C_1 + ...9C_9 = 512$ We have to subtract $9C_0$ and $9C_1$ from both the sides of the equations since we cannot form single digit numbers.&nbsp; =&gt; $9C_2 + 9C_3+...+9C_9=512-1-9$ $9C_2+9C_3+...+9C_9=502$ Therefore, $502$ is the right answer.&nbsp;

Let 'ab' be the two digit number. Where b $\neq$ 0. We will get number 'ba' after interchanging its digit.&nbsp; It is given that 10a+b &gt; 3*(10b + a) 7a &gt; 29b If b = 1, then a = {5, 6, 7, 8, 9} If b = 2, then a = {9} If b = 3, then no value of 'a' is possible. Hence, we can say that there are a total of 6 such numbers.

In a tournament, there are 43 junior level and 51 senior level participants. Let 'n' be the number of girls on junior level. It is given that&nbsp;the number of girl versus girl matches in junior level is 153. $\Rightarrow$ nC2 = 153 $\Rightarrow$ n(n-1)/2 = 153 $\Rightarrow$ n(n-1) = 306 =&gt; n$^{2}$-n-306 = 0 =&gt; (n+17)(n-18)=0 =&gt; n=18&nbsp; (rejecting n=-17) Therefore, number of boys on junior level = 43 - 18 = 25.&nbsp; Let 'm' be the number of boys on senior level. It is given that&nbsp;the number of boy versus boy matches in senior level is 276. $\Rightarrow$ mC2 = 276 $\Rightarrow$ m = 24 Therefore, number of girls on&nbsp;senior level = 51 - 24 = 27.&nbsp; Hence, the number of matches a boy plays against a girl&nbsp; = 18*25+24*27 = 1098

The total number of given points are 11. (10 on circumference and 1 is the center) So total possible triangles = 11C3 = 165. However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160

We have been given that a + b + c + d = 7 Total ways of distributing 7 things among 4 people so that each one gets at least one = $^{n-1}C_{r-1}$ = 6C3 = 20 Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 - 4 = 16

According to the question each column, each row, and each of the two diagonals of the 3X3 matrix add up to the same value. This value must be 15. Let us consider the matrix as shown below:&nbsp; <figure><img class="img-responsive" data-image="grid1.PNG" src="https://cracku.in/media/uploads/grid1_OsZjfqF.PNG"></figure> Now we'll try substituting values from 1 to 9 in the exact middle grid shown as 'x'. If x = 1 or 3, then the value in the left bottom grid will be more than 9 which is not possible. x cannot be equal to 2. If x = 4, value in the left bottom grid will be 9. But then addition of first column will come out to be more than 15. Hence, not possible. If x=5, we get the grid as shown below: <figure><img class="img-responsive" data-image="grid2.PNG" src="https://cracku.in/media/uploads/grid2.PNG"></figure> Hence, for x = 5 all conditions are satisfied. We see that the bottom middle entry is 3.&nbsp; Hence, 3 is the correct answer.&nbsp;

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens Now these two pens have to be divided between three persons so that each person can get zero pens = $^{2+3-1}C_{3-1}$ = $^4C_2$&nbsp; = 6

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as Case I: 2, 3, 4, 6 Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways. Hence total number of ways = 3*6 = 18 Case II: 0, 2, 3, 4 case II a: 0 is in the units place =&gt; 3! = 6 ways case II b: 0 is not in the units place =&gt; units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 * 2 * 2 = 8 Total number of ways in this case = 6 + 8 = 14 ways. Case III: 0, 2, 4, 6 case III a: 0 is in the units place =&gt; 3! = 6 ways case II b: 0 is not in the units place =&gt; units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 * 2 * 2 = 12 Total number of ways in this case = 6 + 12= 18 ways. Hence the total number of ways = 18 + 14&nbsp; + 18 = 50 ways

We have to essentially look at numbers between 1000 and 4000 (including both). The first digit can be either 1 or 2 or 3. The second digit can be any of the five numbers. The third digit can be any of the five numbers. The fourth digit can also be any of the five numbers. So, total is 3*5*5*5 = 375. However, we have ignored the number 4000 in this calculation and hence the total is 375+1=376

The power is 20. 20 has to be divided among a, b and c. This can be done in $^{20+3-1}C_{3-1}$ = $^{22}C_2$ = 231 Option a) is the correct answer.

CAT Probability, Combinatorics

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Instructions

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Free CAT Quant Questions

CAT Probability, Combinatorics

Question 1.

Question 2.

<div><p>The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is</p></div>

Question 3.

<div><p>The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is</p></div>

Question 4.

<div><p>The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is</p></div>

Question 5.

<div><p>The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is</p></div>

Question 6.

<div><p>How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?</p></div>

Question 7.

<div><p>The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is</p></div>

Question 8.

<div><p>A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is</p></div>

Question 9.

<div><p>With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is&nbsp;either to (x, y+1) or to (x+1, y), is</p></div>

Question 10.

Question 11.

<div>How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?</div>

Question 12.

<div>How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?</div>

Question 13.

Question 14.

<div>Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?</div>

Question 15.

<div>In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?</div>

Instructions

<div>Instructions <br />For the following questions answer them individually</div>

Question 16.

Question 17.

<div>In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?</div>

Question 18.

<div>How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?</div>

Question 19.

<div>How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?</div>

Question 20.

<div>What is the number of distinct terms in the expansion of $(a + b + c)^{20}$?</div>

Free CAT Quant Questions