Modern Math - Permutation & Combination - Previous Year CAT/MBA Questions
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P, Q, R and S are four towns. One can travel between P and Q along 3 direct paths, between Q and S along 4 direct paths, and between P and R along 4 direct paths. There is no direct path between P and S, while there are few direct paths between Q and R, and between R and S. One can travel from P to S either via Q, or via R, or via Q followed by R, respectively, in exactly 62 possible ways. One can also travel from Q to R either directly, or via P, or via S, in exactly 27 possible ways. Then, the number of direct paths between Q and R is
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
2-digit such numbers: __ __
Ten’s digit can be filled in 9 ways (i.e., 1, 2, 3, …, 9)
Unit’s digit can be filled in 9 ways (i.e., any of the 10 single digits except the one at ten’s place)
⇒ Total such 2-digit numbers = 9 × 9 = 81
3-digit such numbers: __ __ __
Hundred’s digit can be filled in 9 ways (i.e., 1, 2, 3, …, 9)
Ten’s digit can be filled in 9 ways (i.e., any of the 10 single digits except the one at hundred’s place)
Unit’s digit can be filled in 8 ways (i.e., any of the 10 single digits except the ones at hundred’s and ten’s place)
⇒ Total such 3-digit numbers = 9 × 9 × 8 = 648
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
Case 1: Number of 4-digit numbers
__ __ __ __
Here,
unit’s digit can be either 2, 3, 4 or 5 i.e., 4 ways,
ten’s digit can be chosen from remaining 5 numbers in 5 ways,
hundred’s digit can be chosen from remaining 4 numbers in 4 ways.
thousand’s digit can be chosen from remaining 3 numbers in 3 ways.
⇒ Total such 4-digit numbers = 4 × 5 × 4 × 3 = 240
Case 2: Number of 5-digit numbers
__ __ __ __ __
Here,
unit’s digit can be either 1, 2, 3, 4 or 5 i.e., 5 ways,
ten’s digit can be chosen from remaining 5 numbers in 5 ways,
hundred’s digit can be chosen from remaining 4 numbers in 4 ways.
thousand’s digit can be chosen from remaining 3 numbers in 3 ways.
ten thousand’s digit can be chosen from remaining 2 numbers in 2 ways.
⇒ Total such 5-digit numbers = 5 × 5 × 4 × 3 × 2 = 600
Case 2: Number of 6-digit numbers
__ __ __ __ __
Here,
unit’s digit can be either 1, 2, 3, 4 or 5 i.e., 5 ways,
ten’s digit can be chosen from remaining 5 numbers in 5 ways,
hundred’s digit can be chosen from remaining 4 numbers in 4 ways.
thousand’s digit can be chosen from remaining 3 numbers in 3 ways.
ten thousand’s digit can be chosen from remaining 1 number in 1 way.
One lakh’s digit
⇒ Total such 6-digit numbers = 5 × 5 × 4 × 3 × 2 × 1 = 600
8. Let us consider the sum of unit’s digit of all such numbers.
__ __ __ x
If 1 occupies the units digit, number of such numbers = 3! = 6
Sum of all the units digit of such numbers = 6 × 1 = 6
If 2 occupies the units digit, number of such numbers = 3!/2! = 3
Sum of all the units digit of such numbers = 3 × 2 = 6
If 4 occupies the units digit, number of such numbers = 3!/2! = 3
Sum of all the units digit of such numbers = 3 × 4 = 12
∴ Sum of unit’s digits of all possible numbers = 6 + 6 + 12 = 24
Similarly,
Sum of ten’s digits of all possible numbers = 24 × 10 = 240
Sum of hundred’s digits of all possible numbers = 24 × 100 = 2400
Sum of thousand’s digits of all possible numbers = 24 × 1000 = 24000
⇒ Sum of all such numbers = 24 + 240 + 2400 + 24000 = 26664
Also, number of such numbers = 4!/2! = 12
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
Out of the seven integers given, 3 and 5 should always be selected.
Case 1: Exactly 1 of 7 or 8 is selected.
Exactly one of 7 or 8 can be selected in 2 ways.
Now we have already selected three integers.
For each of the remaining 4 integers, it may be selected or it may not be selected.
∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16
∴ Total number of ways = 2 × 16 = 32
Case 2: None of 7 and 8 is selected.
So far we have selected only two integer.
For each of the remaining 4 integers, it may be selected or it may not be selected.
∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16
But this includes one way when no integer is selected.
∴ Number of ways of selecting at least one integer = 16 – 1 = 15
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
We know that number of distributing n identical objects amongst r different groups = n+r-1Cr-1
Distributing balloons
Since each child gets at least 4 balloons, we will first give 4 balloons to each of the 3 children.
Now we have 3 identical balloons to be distributed to 3 children.
Number of ways of doing this = 3+3-1C3-1 = 10 ways
Distributing pencils
Since each child gets at least 1 pencil, we will first give 1 pencil to each of the 3 children.
Now we have 3 identical pencils to be distributed to 3 children.
Number of ways of doing this = 3+3-1C3-1 = 10 ways
Distributing erasers
We have 3 identical erasers to be distributed to 3 children.
Number of ways of doing this = 3+3-1C3-1 = 10 ways
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
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 either to (x, y + 1) or to (x + 1, y), is
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?
In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is
Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. 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?
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?
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?
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?
For the 4 digit number to be divisible by 6, the sum of the digits of the 4 digit number has to be a multiple of 3 and the unit (right most) digit has to be an even number.
The combination of 4 digits that add upto a multiple of 3 are (0, 2, 3, 4), (0, 2, 4, 6) and (2, 3, 4, 6)
Case 1: Combination (0, 2, 3, 4)
For this combination the required number can start with 2, 3 or 4.
First digit 2: The last digit can be 0 or 4. The 2nd and 3rd digit from the left can be one 0/4 or 3. So a total of 4 possibilities.
First digit 4: Here too the number of possibilities is 4 as the 2nd and 3rd digit from the left can be one of 0/2 or 3.
First digit 3: Digits 0, 2 and 4 can be in any order as the 2nd, 3rd and 4th digit from the left. So a total of 3! or 6 possibilities.
So number of possibilities with combination (0, 2, 3, 4) = 4 + 4 + 6 = 14
Case 2: Combination (0, 2, 4, 6)
The first digit can be anyone of 2, 4 or 6 i.e., 3 possibilities. The 2nd digit can be anyone from 0/2/4/6 but apart from the one used in the left most digit i.e., 3 possibilities. The 3rd digit from the left will be any one from 0/2/4/6 but apart from the 2 digits used in the leftmost or 2nd leftmost digit i.e., 2 possibilities. The extreme right digit will be the last digit after 3 out of these 4 digits are used in the first 3 digits from the left.
So total number of possibilities = 3 × 3 × 2 = 18
Case 3: Combination (2, 3, 4, 6)
The units digit has to be one amongst 2, 4 or 6 i.e., 3 possibilities. So first 3 digits from left will be 3 and two amongst 2/4/6 (i.e., except that digit which is not chosen as the right most digit) and can be arranged in 3! or 6 ways
So number of possibilities with combination (2, 3, 4, 6) is 6 × 3 = 18
So number of 4 digits numbers that can be formed using digits 0, 2, 3, 4 and 6 is 14 + 18 + 18 = 50
The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.
Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is
We can find the number of shortest possible paths from A to E either by trial and error or by using combinations.
Note that to travel from A to E, we have to take 2 roads to the right and 2 roads downwards (in the diagram) in order that we follow the shortest path. In other words, we have to use 2 + 2 = 4 roads, out of which 2 are towards right and 2 are downwards.
This is equivalent to selecting 2 things (roads towards right) out of 4 things (roads). (The remaining two roads will be downwards.)
The number of ways of doing this is 4C2 = 4!/(2!×2!) = 6
∴ From point A to E, there are 6 ways to reach with the minimum distance travelled.
Here E to F is the shortest distance because the third side of a triangle is always less than the sum of the other two sides.
From point F to B, there are 6C4 = 6!/(4!×2!) = 15 ways to reach with the minimum distance travelled.
∴ There are 15 × 6 = 90 shortest paths that Neelam can choose.
Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is
Answer the next 2 questions based on the information given below.
Let S be the set of all pairs (i, j) where 1 ≤ i < j ≤ n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1, 2) and (2, 3) are also friends, but (1, 4) and (2, 3) are enemies.
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