IISER Mathematics - Permutations and Combinations
Exam Duration: 45 Mins Total Questions : 30
Ten different letters of an alphabet are given. Words with 5 letters are formed from these given letters. Then the number of words which have at least one letter repeated is
- (a)
69,760
- (b)
30,240
- (c)
99,748
- (d)
NONE OF THESE
Number of 5 letter words (with the condition that a letter can be repeated) = 10s. Again, number of words using 5 different letters is 10P5. Therefore, required number of letters = Total number of words - Total number of words in which no letter is repeated = 105- 10P5= 69760.
The number of ways in which 20 rupees can be distributed among five persons such that no person receives less than 3 rupees is
- (a)
125
- (b)
126
- (c)
127
- (d)
NONE OF THESE
Number of ways
= coefficient of x20 in
(x3+x4+x5.x20.)5
=coefficient of x20 in
x15 (1+ x + x2+ x15)5
=coefficient of x5 in \({ \left( \frac { 1-x^{ 17 } }{ 1-x } \right) }^{ 5 }\)
= coefficient of x5 in (l - x)-5 (1 - x17)5
= coefficient of x in (l - x)-5
(\(\because \)(1 - x17)5 does not give coeff. of x5)
= coefficient of x5in
[1 +5x + 15x2 + 35x3 + 70x4 + 126x5+ ..]
=126
If the best and the worst paper never appear together, then six examination papers can be arranged in
- (a)
120 ways
- (b)
480 ways
- (c)
240 ways
- (d)
NONE OF THESE
Total numbers of ways of arranging 6 papers = 6! = 720. Let the best and the worst papers be treated as one unit, then the number of ways in which these papers, in which best and worst papers are together = 5!. The best and the worst papers can interchange their positions in 2 ways.
Number of ways in which the best and the worst papers are together = 2 x 5! = 240.
Required number of ways = 720 - 240 = 480.
There are 10 true-false questions in an examination. Then these questions can be answered in
- (a)
20 ways
- (b)
100 ways
- (c)
512 ways
- (d)
1024 ways
Each question can be answered as true or false i.e. in 2 ways.
Total number of ways
=2x2x2x2x2x2x2x2x2x2
=210=1024 ways.
If all the letters of the word AGAIN be arranged as in a dictionary, then the rank of the word NAAIG, is
- (a)
47
- (b)
48
- (c)
49
- (d)
50
Starting with A, the letters GAIN can be arranged in = 4! ways = 24 ways
The number of words beginning with A = 24
Starting with G, then letters AAIN can be arranged in \(\frac{4!}{2!}\) = 12 ways
The number of words beginning with G = 12
Similarly, the number of words beginning with I= \(\frac{4!}{2!}\) = 12
The number of words beginning with these 3 letters = 24 + 12 + 12 = 48
Now, 49th word is NAAGI. and 50th word is NAAIG
Thus, rank of the word NAAIG is 50
Hence, the correct alternative is (d).
The total number of permutations of n(>1) different things taken not more than r at a time, when each thing may be repeated any number of times is
- (a)
\({n(n^n-1)\over{n-1}}\)
- (b)
\({n^r-1\over{n-1}}\)
- (c)
\({n(n^r-1)}\over{n-1}\)
- (d)
None of these
When we arrange one thing at a time, then the number of possible permutations is n. When we arrange them two at a time, the number of possible permutations are nXn=n2 and so on.Thus the total number of permutations are
\(n+n^2+...+n^r={n(n^r-1)\over(n-1)}\)
A student is allowed to select atmost n books from a collection of (2n+1) books. If the number of ways in which he can select atleast one book is 63, then is equal to
- (a)
3
- (b)
4
- (c)
6
- (d)
5
He can select t, 2, or n books.
The number of ways to select atleast one book is
2n+1C1+ 2n+1C2+.....+2n+1Cn
=\(\frac{1}{2}\)(2n+1C1+ 2n+1C2+.....+2n+1Cn+2n+1Cn+1+.....+2n+1C2n)
=\(\frac{1}{2}\)(22n+1-2n+1C0-2n+1C2n+1)
=22n-1=63 [given]
\(\Rightarrow\) 22n=64=26\(\Rightarrow\) n=3
There are 4 balls of different colours and 4 boxes of colours same as those of the balls. The number of ways in which the balls, one in each box could be placed such that a ball does not go to a box of its own colour, is
- (a)
5
- (b)
6
- (c)
9
- (d)
12
Number of ways = \(4!{({1\over2!}-{1\over3!}+{1\over4!})}\)
=12-4+1=9
There are m points on one straight line AB and n points on another straight line AC, none of them being A. How many triangles can be formed with these points as vertices, if point A is also included?
- (a)
\({mn\over2}(m+n-2)\)
- (b)
m+1C2 X nC1 + m-1C1 X nC2
- (c)
\({mn\over 2}{(m+n)}\)
- (d)
None of the above
A triangle can be constructed in two ways
(i)By taking two points on AB and one point on AC.
(ii)By taking two points on AC and one point on AB.
If point A is included, then total number of triangles= m+1C2XnC1+mC1XnC2= \({mn\over 2}{(m+n)}\)
The remainder obtained, when 1! + 2! + 3! + ... + 175! is divided by 15 is
- (a)
5
- (b)
0
- (c)
3
- (d)
8
\(\because\) \(\downharpoonright 5,\downharpoonright 6,\downharpoonright 7,\downharpoonright 8,...,\downharpoonright 175\) each multiple of 15
and \(\downharpoonright 1+\downharpoonright 2+\downharpoonright 3+\downharpoonright 4=33\)
Hence, required remainder = 3.
The sides AB, BC and CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these interior points as vertices, is
- (a)
205
- (b)
208
- (c)
220
- (d)
380
12C3 - 3C3 - 4C3 - 5C3 = 205
The number of triangles whose vertices arc the vertices of an octagon but none of whose sides happen to come from the octagon is
- (a)
16
- (b)
28
- (c)
56
- (d)
70
Number of triangles = total number of triangles
- number of triangles having one side common
- number of triangles having two sides common
= 8C3 - 8 X 4 - 8
= 56 - 32 -8 = 16
There are n points in a plane of which no three are in a straight line except 'm' which are all in a straight line. Then the number of different quadrilaterals, that can be formed with the given points as vertices. is
- (a)
nC4 - mC3n-m+1C1-mC4
- (b)
nC4 - mC3n-mC1+mC4
- (c)
nC4 - mC3n-mC1-mC4
- (d)
nC4 - nC3.mC1
Number of quadrilaterals = total number of quadrilaterals
- choosing 3 points from m and 1 from remaining points
- choosing 4 points from m points
= nC4 - mC3 (n-mC1)-mC4
The number of divisors a number 38808 can have, excluding 1 and the number itself is
- (a)
70
- (b)
72
- (c)
71
- (d)
none of these
Factorizing the given number, we have
38808 = 23.32.72.11
Therefore the total number of divisors
= ( 3 + 1 ) ( 2 + 1 ) ( 2 + 1 ) ( 1 + 1 ) - 1 = 71
But this includes the division by the number itself.
Hence, the required number of divisors = 71 - 1 = 70.
The maximum number of different permutations of 4 letters of the word EARTHQUAKE is
- (a)
1045
- (b)
2190
- (c)
4380
- (d)
2348
Here 2E, 2A, 1T, 1H, 1Q, 1U, 1K
Maximum number of permutations = coefficient of x4 in \(\downharpoonright 4{ \left( 1+x \right) }^{ 6 }\left( 1+\frac { x }{ \downharpoonright 1 } +\frac { { x }^{ 2 } }{ \downharpoonright 2 } \right) \)
= coefficient of x4 in 6 ( 1 + x )6 [ ( 1 + x )2 + 1 ]2
= coeffcient of x in 6 {( 1 + x )10 + 2 ( 1 + x )8 + ( 1 + x )6}
=6.{ 10C4 + 2.8C4 + 6C4 }
\(=6.\left\{ \frac { 10.9.8.7 }{ 1.2.3.4 } +2.\frac { 8.7.6.5 }{ 1.2.3.4 } +15 \right\} \)
= 2190
The number of ways in which we can choose 2 distinct integers from 1 to 100 such that difference between them is at most 10 is
- (a)
10C2
- (b)
72
- (c)
100C2 - 90C2
- (d)
none of these
\(\because\) \(| x - y|\le 10\) Total ways = 100C2
and is \(|x-y|\ge10\) then, \(x-y\le10\) and \(x - y \ge 10\)
ie, \(x \le y-10\) and \(x \ge y +10\)
ie, take tow values out of 90 or 90C2
\(\therefore\) Required number of ways = 100C2 - 90C2
Consider seven digit number x1, x2, ....., x7, where x1, x2, .....,x7 \(\neq\) 0 having the property that x4 is the greatest digit and digits towards the left and right of x4 are in decreasing order. Then total number of such numbers in which all digits are distinct is
- (a)
9C7 . 6C3
- (b)
9C6 . 5C3
- (c)
10C7 . 6C3
- (d)
9C2 . 6C3
Required number =number of selections of one or more out of three 25 paise coins and two 50 paise coins = 4 x 3-1 = 11 = 12p1 -1
A letter lock consists of three rings marked with 15 different letters. If N denotes the number of ways in which it is possible to make unsuccessful attempts to open the lock, then
- (a)
482 | N
- (b)
N is product of three distinct prime numbers
- (c)
N is product of four distinct prime numbers
- (d)
N is product of two distinct prime numbers
The total number of attempts that can be made to open the lock is (15)3. Out of these, there is just are attempt in which lock will open
Then, N = (15)3 -1 = (15 - 1) ((15)2 + (15) + 1)
= 14 (225 + 15 + 1)
= 2x 7 x 241
Clearly N is divisible by 482 ie, 482|N and N is product of three distinct prime numbers.
Let N denote the greatest number of points in which m straight lines and n circles intersect, then
- (a)
m | (N - mC2 - nP2)
- (b)
n | ( N - mC2 - nP2)
- (c)
N - mC2 - nP2 is an ever integer
- (d)
N - mC2 - nP2 is an odd integer
m straight lines can intersect is at most mC2 points and n circles can intersect in at most 2 X nC2 = n (n - 1) = nP2 points.
11 straight lines and m circles can intersect is at most 2 x m x 11 = 2 mn points
Hence, N = mC2 + nP2 + 2 mn
or ( N - mC2 - nP2 ) = 2mn = is an even integer
Clearly m | (N - mC2 - nP2) and n| (N - mC2 - nP2)
Suppose a lot of n objects contains n 1 objects of one kind, n 2 objects of second kind, n3 objects of third kind,....., n k objects of kth kind. Such that n1 + n2 + n3 + ... + n k = n, then the number of possible arrangements/permutations of r objects out of this lot is the coefficient of x r in the expansion of \(r!\Pi \left( \overset { { n }_{ 1 } }{ \underset { \lambda =0 }{ \Sigma } } \frac { { x }^{ \lambda } }{ \lambda ! } \right) \)
If n1 + n2 + n3 + ... + nk = r, then number of permutations must be
- (a)
nCr
- (b)
nPr
- (c)
( k + r )!
- (d)
\(\frac {r!}{{n_1}!{n_2}!...{n_k}!}\)
\(\because\) n1 + n2 + n3 + ... + nk = r
\(\Rightarrow\) n = r
Number of permutations = number of permutations of r objects in a line of which n1 are of one kind, n2 of second kind, n3 of third kind, ..., nk of kth kind \(= \frac {r!}{{n_1}!{n_2}!...{n_k}!}\)
Let f(n)f(n) denotes the number of different ways the positive integer n can be expressed as the sum of 1's and 2's. For example
f(4)=5f(4)=5
ie, 4=1+1+1+1
=1+1+2
=1+2+1
=2+1+1
=2+2
The number of solutions of the equation f(n) = n + 1, where n \(\in\) N is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Only f(4) = 5 (of above discussion)
ie, f(n) = n + 1.
Different words are being formed by arranging the letters of the word 'ARRANGE'. All the words obtained are written in the form of a dictionary.
The number of words in which at least one vowel is in between two consonant is
- (a)
18
- (b)
36
- (c)
624
- (d)
836
The consonants are N, G, R, R, and vowels are A, A, E X A X A X E X
Hence number of words at least one vowel is in between two consonants is \(\frac {4!}{2!}\times \frac{3!}{2!}=12\times 3 =36\)
In a class, there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl to represent the class for a function. In how many ways can the teacher make this selection?
- (a)
378
- (b)
377
- (c)
375
- (d)
379
Here the teacher is to perform two operations:
(i) Selecting a boy from among the 27 boys and
(ii) Selecting a girl from among 14 girls.
The first of these can be done in 27 ways and second can be performed in 14ways. By the fundamental principle of • counting, the required number of ways is 27 x 14 = 378.
In an examination, there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answers correct is
- (a)
11
- (b)
12
- (c)
27
- (d)
63
There are three multiple choice questions, each has four possible answers. Therefore, the total number of possible answers will be 4 x 4 x 4 = 64. Out of these possible answers only one will be correct and hence the number of ways in which a student can fail to get all correct answers is 64 - 1 = 63.
How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
- (a)
3024
- (b)
3026
- (c)
3040
- (d)
3014
There will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4 digit numbers
=\({ { 9 }_{ P } }_{ 4 }=\cfrac { 9! }{ \left( 9-4 \right) ! } =\cfrac { 9! }{ 5! } =9\times 8\times 7\times 6=3024\)
The sum of all the five digit numbers formed with the 1,2,3,4,5 taken all at time is
- (a)
15(5!)
- (b)
3999960
- (c)
3990000
- (d)
none of these
Sum of unit's place = 4! (l + 2 + 3 + 4 + 5) = 24 xIS = 360
[Since 1 will come 24 times in unit place and so the other digits if 5 digit number is formed with digits 1, 2, 3, 4, 5 taken all at a time]
So the sum in ten's, hundred's, thousand's and ten thousand's places is
= 360 x 10000 + 360 x 1000 + 360 x 100 + 360 x 10 + 360
= 3999960.
Find the number of arrangements of the letters of the word INDEPENDENCE when words begin with I and end in P.
- (a)
12400
- (b)
12420
- (c)
12600
- (d)
12620
Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters.
Hence,the required number of arrangements
= \(\cfrac { 10! }{ 3!2!4! } =12600\)
If nC9=nC8, find nC17
- (a)
1
- (b)
2
- (c)
0
- (d)
3
nCp=nCk\(\Rightarrow\)n = p + k
\(\therefore\) nC9=nC8\(\Rightarrow\)n = 9 + 8 = 17
Now 17C17=1
Find the number of ways of choosing 4 cards from a pack of 52 playing cards when cards are of the same colour.
- (a)
29900
- (b)
29925
- (c)
29910
- (d)
29920
4 red cards can be selected out of 26 red cards in 26C4 ways. 4 black cards can be selected out of 26 black cards in 26C4 ways.
Therefore,the required number of ways= 26C4+26C4
= \(2\times \cfrac { 26! }{ 4!22! } =29900\)
There are 25 points in a plane, of which 1 are on the same line. Of the rest, no three are collinear and no two are collinear with any one of the first ten points. The number of different straight lines that can be formed by joining these point is
- (a)
256
- (b)
106
- (c)
255
- (d)
2105
Out of 25 given points, 10 are collinear and hence they form only one straight line. Out of rest of the 15 points, we have 15C2 straight lines and anyone point out of these 15 points with anyone of 10 collinear points forms a straight line.
Hence, total straight lines formed
=15C2+15C1 X 10C1+ 1 = 256.