In P&C (Permutation and Combination), it is more important to understand the problem than to know how to solve it. This is because the task is to figure out how the problem is to be solved, while the calculation part is too easy to worry about. Even if you do not know the formula or property, you can still get the answer you need by being creative, clever, and artistic. As a business manager or team leader, it is your job to set up events and choose the best people and tools, so why do not you start right now.
Since it is a very logical part, it is one of the ones that people who set the questions for competitive tests like to use the most. About 1-3 questions from this chapter will come up in almost every CAT session/slot. Other tests, like the XAT and IIFT, also ask about two to four problems from this chapter.
Counting
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If one operation can be done in m ways and a second operation can be done in n ways for each way the first operation can be done, then the two operations can be done together in (m x n) ways. If a third operation can be done in p ways after two operations have been done in any of the (m x n) ways, then the three operations can be done together in (m x n x p) ways, and so on.
Permutations
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You take all or some of a set of finite things and arrange them in different ways. Each of these different ways is called a permutation. In Permutation, how things are put together is very important.
Permutation means selection and arrangement both, while Combination means only selection.
Types of Permutations
- A permutation is said to be a Linear Permutation if the objects are arranged in a line. A linear permutation is simply called as a permutation.
- A permutation is said to be a Circular Permutation if the objects are arranged in the form of a circle.
- The number of (linear) permutations that can be formed by taking r things at a time from a set of n distinct things is denoted by nPr or P(n, r), for every 1 ≤ r ≤ n.
- nPr =n(n−1)(n−2)(n−3)….....(n−r+1)= \(\frac{n!}{(n-r)!}\)
Permutations of n Different Things
- Number of permutations of n different things taken all at a time = nPn = n!
- No of permutations of n different things taken r at a time= nPr = \(\frac{n!}{(n-r)!}\)
- Number of permutations of n different things taken at Most r at a time= nP + nP + nP +...+ nPr
- Number of permutations of n different things taken at least r at a time = nPr +nPr+1 +nPr+2 +...+nPn
- Number of permutations of n different things taken r at a time, when one particular thing always occurs = r ⋅ ( n −1Pr − 1 )
- Number of permutations of n different things taken r at a time, when one particular thing never occurs = n −1Pr
- Number of permutations of n different things taken r at a time, when k particular things, always occur = (rPk )(n −k Pr − k )
- Number of permutations of n different things taken r at a time, when k particular things never occur = n −k Pr
- Number of permutations of n different things, taken all at a time, when m specified things always come together = m!(n−m+1)!
- Number of permutations of n different things, taken all at a time, when m specified things never come together = n! − [m!(n − m +1)!]
Permutations of n Things Not All Different
- Number of ways to arrange n things when p of them are all the same and the rest are all different = \(\frac{n!}{p!}\)
- Number of ways n things can be put together when p things are the same of one kind, q things are the same of another kind, r things are the same of another kind, and the rest, if any, are different from p, q, and r. = \(\frac{n!}{p!q!r!}\)
- Number of ways n things taken all at once when p1 things are alike of one kind, p2 things are alike of second kind, p3 things are alike of third kind,... pr things are alike of rth kind, and the rest, if any, are different from p1, p2, p3,..., pr, then the number of permutations of n things = \(\frac{n!}{p_1! p_2!........p_r!}\)
Permutations Where Repetitions are Allowed
- Number of permutations of n different things taken exactly r at a time, where each thing can be repeated any number of times in each order = nr
- Number of different things that can happen at any given time, and the number of times each thing can happen in each order. = n+n2 +n3 +...+nr = \(\frac{n(n^r-1)}{(n-1)}\)
- Number of permutations of n different things taken at least r at a time, when each may be repeated any number of times in each arrangement = nr +nr+1 +nr+2 +….+nn = \(\frac{n^r(n^{(n-r+1)}-1)}{n-1}\)
Circular Permutations
- Number of circular permutations of n different things taken all at a time = (n −1)!
- Number of circular permutations of n different things taken all at a time in one direction (either clock-wise or anti-clock-wise) = \(\frac{(n-1)!}{2}\)
- Number of circular permutations of n different things taken r at a time = \(\frac{^np_r}{r}\)
- Number of circular permutations of n different things taken r at a time in one direction (either clock-wise or anti-clock-wise) = \(\frac{^np_r}{2r}\)
Combinations
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number of combinations of n dissimilar things taken r at a time is denoted by nCr or C(n,r) or (nr)
n Cr = \(\frac{n!}{r!\ (n-r)!}\) C denotes the combination and r ≤ n
Essential Properties
- nCr = \(\frac{p_r}{r!}\)
- n C0 = n Cn = 1
- n Cr = n Cn-r ( 0≤r≤n)
- n Cr = n Cy \(\Leftrightarrow\) x+y =n or x = y
- nCr−1 +nCr =n+1Cr
- If n is even, the greatest value of n Cr = nCm where m = n/2
- If n is odd, the greatest value of nCr = nCm where m= \(\frac{(n \mp 1)}{2}\)
- rCr +r+1Cr +r+2Cr +…....+nCr = n+1Cr+1 ; r ≤ n
- nC0 + nC1 +n C2 + nC3 +…...+ nCn = 2n
- nC1 +nC2 +nC3 +…...+nCn = 2n −1
- nC0 +nC2 +nC4 +…....= 2n−1
- nC1 +nC3 +nC5 +….....= 2n−1
Combination of n Different Things
- Number of ways that n different things can be put together when x things always happen = n − xCr − x
- Number of ways n different things can happen at the same time when x things never happen = n − x C r
- Number of ways that n different things can be put together so that x things always happen and y things never happen = n −x−yCr−x
- Number of ways n different things can be put together r at a time when x things can not be put together in any pick = n-xC r −x
- Number of ways of selections of zero or more things from a group of n distinct things = nC0 +nC1 +nC2 +nC3 +...+nCn =2n
- Number of ways of selections of one or more things from a group of n distinct things = nC1 +nC2 +n C3 +...+nCn =2n −1 (nC0 =1)
Combination of n Identical Things
- Number of ways of selections of r things from a group of n identical things = 1.
- Number of ways of selections of at most r things from a group of n identical things = r + 1
- Number of ways of selections of at least r things from a group of n identical things = (n − r) +1
- Number of ways of selections of at least one thing from a group of n identical things = n
- Number of ways of selections of zero or more things from a group of n identical things = n + 1
Combination of n Things Not All Different
- Number of ways of selection of zero or more things out of (p+q+r+...)things,ofwhich parealikeofone kind, q are alike of second kind, r are alike of third kind,andsoon=[(p+1)(q+1)(r+1)...]
- Number of ways of selection of at least one thing out of (p+q+r+...)things,ofwhich parealikeofonekind, q are alike of second kind, r are alike of third kind, and so on =[(p+1)(q+1)(r+1)K]−1
- Number of ways of selection of zero or more things from p identical things of one kind, q identical things of second kind, r identical things of third kind, and from remaining n things which are all different = [( p +1) (q +1) (r +1)2n ]
- Number of ways of selection of at least one thing from p identical things of one kind, q identical things of second kind, r identical things of third kind, and from remaining n things which are all different =[{(p+1)(q+1)(r+1)2n}−1]
- The number of ways of choosing k objects, from
p objects of one kind, q objects of second kind, and so on, is the coefficient of xk in the expansion (1+x+x2 +K+xp)(1+x+x2 +K+xq)... - The number of ways of choosing k objects from
p objects of one kind, q objects of second kind, and so on, such that one object of each kind must be included is the coefficient of xk in the expansion (x+x2 +....+xp)(x+x2 +K+xq) - The number of ways in which r things can be selected from a group of n things of which p are identical is ∑n − p Cr ; if r ≤ p
- The number of ways in which r things can be selected from a group of n things of which p are identical is ∑n−pCr ;if r > p
Combination of Contiguous Things
(A) Linear Combination
- Number of selections of k consecutive things out of n things in a row=(n−k +1)
- Number of selection of k things out of n things in a row, such that no two of the selected objects are next to each other= n−k+1Ck
(B) Circular Combination
Number of selections of k consecutive things out of n things in a circle
n, when k < n
1, when k = n
Previous Year CAT Questions
Ques 1: If 0 < x < 270°, then what is the probability that sin x > cos x ? (CAT 2016)
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Ans:
If n even, i.e. say n = 2 m, then the number of ways is
2 × m ! × m ! , i.e. m odd numbers in alternate places and m even numbers in alternate places.
If n is odd, i.e. say n = 2m + 1, then
Number of ways = m!(m + 1)!
Hence, either 2(m!)2 = 72 or m!(m + 1)! = 72
If 2(m!)2 =72 ⇒m!=6⇒m=3
For m!(m + 1)! = 72, there is no solution.
Hence, m = 3 and n = 2m = 6.
Ques 2: One red flag, three white flags and two blue flags are arranged in a line such that
- No two adjacent flags are of the same colour.
- The flags at the two ends of the line are of different colours.
In how many ways can the flags be arranged? (CAT 2015)
- 7
- 3
- 2
- 6
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Ans: (D)
The required possibilities are given as under w*w*w*or *w*w*w, where *shows the position occupied by 2 blue and 1 red flag.
∴ Total number of permutations = 2\(\frac{3!}{2!}=6\)
Ques 3: From 5 consonants and 4 vowels ......... words can be formed using 3 consonants and 2 vowels? (CAT 2015)
- 6500
- 7200
- 6200
- 5200
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Ans: (B)
From 5 consonants, 3 consonants can be selected in 5 C3 ways. From 4 vowels, 2 vowels can be selected in 4C2
4C2 ways. Now, with every selection, number of ways of arranging 5 letters is 5 P5 .
∴ Total number of words = 5C3 × 4C2 × 5 P5 = 10 x 6 x 5 x 4 x3 x 2 x 1 = 7200
Ques 4: In a chess tournament held this year in kolkata, there were only two women participant among all the members participate in the tournament. Every participant played two games with each other participant. The number of games that men played between them selves proved to exceed by 66, as compared to the number of games the men played with women. How many participant were there in the tournament? (CAT 2013)
- 156
- 13
- 610
- 108
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Ans: (B)
Total number of women participant be x.
Since, every participant played two games with each
other participant.
So, total number of games played among men = 2 x nC2 =
2 x n!/ 2! (n-2)! = n(n – 1)
Number of games played with each women = 2n
Since, each women must have played two games with each men.
∴ Total match played by women =2 ×2n =4n
Now, according to the question,
n (n − 1) − 4n = 66 n2 − n −4n =66
n2 − 5n − 66 = 0
n2 − 11n + 6n − 66 = 0
(n − 11) (n + 6) = 0
n = 11, − 6
(Q n > 0 , so – 6 is not possible)
∴ Number of total participant = 11 + 2 = 13
Ques 5: Letters of the word ‘‘ATTRACT’’ are written on cards and are kept on a table. Manish is asked to lift three cards at a time, write all possible combinations of the three letters on a piece of paper and then replace the three cards. The exercise ends when all possible combinations of letters are exhausted. Then, he is asked to strike out all words in his list, which look the same when seen in a mirror. How many words is he left with? (CAT 2012)
- 40
- 20
- 30
- None of these
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Ans: (A)
To find all the 3 letters words possible, we consider 4 cases
Number of words using 3 T’ s→1
- Number of words using 2 T’s→3 × 3 = 9, since the 3rd letter may be anyone of A, C, R and can be placed in anyone of 3 positions eg., CTT, TCT, TTC.
- Number of words using 2 A’s → 3 ×3 =9
- Number of words with all 3 letters distinct → 24
Total=24+9+9+1=43words
3 words TTT, ATA, TAT are striken out as they look the same in a mirror.
∴ 43 −3 =40 words.
Ques 6: A student is asked to form numbers between 3000 and 9000 with digits 2, 3, 5, 7 and 9. If no digit is to be repeated, in how many ways can the student do so? (CAT 2012)
- 24
- 120
- 60
- 72
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Ans: (D)
The first digit can be chosen from 3, 5, 7 in 3 ways. Having chosen the first digit, the remaining three digits can be chosen from the remaining four numbers in 4P3 = 24 ways.
∴Total number of ways = 3 ×24 = 72
Ques 7: Rajat draws a 10 × 10 grid on the ground such that there are 100 identical squares numbered 1 to 100. If he has to place two identical stones on any two separate squares in the grid, how many distinct ways are possible? (CAT 2011)
- 2475
- 4950
- 9900
- 1000
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Ans: (B)
Two identical stones are to be placed on any two separate squares in the grid.
∴ r = 2
Total number of squares = 100
∴Total number of stones, n = 100
The number of ways of selecting two things out of 100 is nCr = 100C2 = 4950
Ques 8: 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? (CAT 2010)
- 499
- (b) 500
- (c) 375
- (d) 376
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Ans: (D)
The first number in the set is 1000, which has four digits.
The highest number in the series is 4000, which is also the only four-digit number that begins with 4.
Each of the four-digit numbers other than 4000 can have either a 1 or a 2 or a 3 in the thousand-spot on the left.
The next three digits, which are the hundreds, tens, and units places, can be any of the five numbers 0, 1, 2, 3, or 4.
So, there are 3555, or 375, numbers between 1000 and 3999.
With 4000, there will be a total of 376 of these numbers.
Ques 9: There are four boxes. Each box contains two balls: one red and one blue. You draw one ball from each of the four boxes .What is the probability of drawing at least one red ball? (CAT 2010)
- 1/2
- 1/4
- 1/16
- 15/16
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Ans: (D)
probability of at least on red ball = 1 − (Probability of all blue balls)
the probability of drawing one blue ball is (1/2)
the probability of drawing all four blue balls will be = \((\frac{1}{2})^4 = \frac{1}{16}\)
Probability of one blue ball = 1-(1/16)= (15/16)
Ques 10: A garland is to be made from six different flowers and a large pendant which has two different faces. In how many ways can the garland be made? (CAT 2009)
- 240
- 600
- 720
- None of these
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Ans: (C)
Six different flowers and a large pendant are 7 different things that are to be arranged in a circular manner, which can be done in (7 −1)! = 6! = 720 ways.
Ques 11: Five persons entered the lift cabin on the ground floor of an seven storied building. Suppose that each of them independently and with equal probability, can leave the cabin at any floor beginning with the first. What will be the probability of all the five persons leaving at different floors? (CAT 2009)
- 0.02
- 0.15
- 0.37
- 0.38
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Ans: (B)
Besides the ground floor, there are seven floors. The total number of ways in which each of the five persons can leave the cabin at any of the 7 floors = 75
And, the favourable number of ways, i.e. the number of ways in the which the 5 persons leave at different floors is 7P5
Probability = (7P5/75) = 0.15
Ques 12: In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is __ (CAT 2005)
- 200
- 216
- 235
- 256
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Ans: (A)
Let there be m boys and n girls.
nC2 = 45 = n(n – 2)/2
n(n-1)= 90
n=10
m (m-1)/2= 190
m(m-1)= 380
m=20
Ques 13: For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive? x2 −y2 = 0, (x−k)2 +y2 = 1. (CAT 2005)
- 2
- 0
- 2
- −2
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Ans: (C)
y2=x2
2x2 − 2kx + k2 − 1 = 0
D=0
⇒ 4k2 =8k2 −8
⇒ 4k2 =8
⇒ k=2
Ques 14: In the figure, given below the lines represent one-way roads allowing travel only Northwards or only Westwards. Along how many distinct routes can a car reach point B from point A? (CAT 2004)
- 15
- 56
- 120
- 336
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Ans: (B)
Any route from A to B consists of 3 + 5 = 8 segments, where the car can move only 5 segments to the West and only 3 segments to the North.
The number of distinct routes is equal to the number of ways of choosing 3 out of the 8 segments along which the car can go North or choosing 5 segments along which the car can go West.
Therefore, the number of distinct routes from A to B is 8C3 = \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\)
Ques 15: A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is (CAT 2004)
- 12 × 81
- 20 ×125
- 16 × 192
- 24 × 216
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Ans: (A)
Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colour of the second stripe).
Similarly, there are 3 ways to colour each of the remaining stripes.
∴The number of ways the flag can be coloured is 54
4(3)5 = (12)(34 ).
How to approach Permutation and Combination Questions in CAT
- Understand the question first and then think of applying formulae to get the answers
- Go through the formulae for easily solve the questions
- Practice as much analytical question to get a hang of all those situation to solve questions using permutation and combination.
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