PERMUTATIONS AND COMBINATIONS - 3 (Simple Applications on nPr and nCr)

Important Results

1. Total number of selections of one or more objects from n different objects =nC1+nC2+.+nCn=2n1

2. Total number of selections of any number of things from n identical things

={(n+1); when selection of zero things is allowed n; when at least one thing is to be selected. 

3. Total number of selections from p like things, q like things of another type and r distinct things

={(p+1)(q+1)2r1 (if at least one thing to be selected) (p+1)(q+1)2r2( if none or all cannot be selected) 

4. Total number of selections of r things from n different things when each thing can be repeated unlimited number of times n+r1Cr1

5. Total number of ways to divide n identical things among r persons =n+1Cr1

Results on distribution

6. Distribution of n things to r boxes

Given Condition Number of ways
n distinct things Empty boxes are allowed rn
r distinct boxes Empty boxes are not allowed coefficient of xn in n!(ex1)r
n identical things Empty boxes are allowed n+r1Cr1
r distinct boxes Empty boxes are not allowed n1Cr1

Division of items into groups

7. (i) Groups of unequal size.

  • Number of ways in which (m+n+p) items can be divided into unequal groups containing m,n,p items is (m+n+p) ! m!n!p!
  • Number of ways to distribute (m+n+p) items among 3 persons in the group containing m,n&p items is (m+n+p)!m!n!p!3 !

(ii) Groups of equal size

  • Number of ways in which (mn) different items can be divided equally into m groups each containing n objects

={(mn)!(n!)mm!; if order of gorups is not impor tan t(mn)!(n!)m; if order of groups is impor tan t

Note

(i) If there are m items of one kind, n items of another kind, then the number of ways of choosing r items out of these = coefficient of xr in (1+x+x2+.

(ii) If there are m items of one kind n items of another kind, then the number of ways of choosing r items such that at least one item of each kind is included

= coefficient of xr in (x+x2+..+xm)(x+x2+.+xn)

8. Results related with points, Lines, Rectangle, Polygon, Circle, etc.

(i) If these are n points in the plane, number of line segments nC2

(ii) Number of points n, then the number of triangles nC3

(iii) Number of diagonals in a regular polygon having n sides =nC2n

(iv) Number of parallelograms when a parallelogram is cut by two sets of m lines parallel to its sides =m+2C2m+2C2

(v) { Number of rectan gles =r=1nr3 Number of squares =r=1nr2

(vi) { Number of rec tan gles =m+1C2n+1C2=mn4(m+1)(n+1) Number of squares =r=1n(mr+1)(nr+1)

(vii) Maximum number of parts in which a plane can be divided by n straight lines =1+r=1nr

(viii) Maximum number of points of intersection of n straight Lines =1×nC2

(ix) Maximum number of points of intersection of n circles =2×nC2

(x) Maximum number of points of intersection of n parabolas =4×nC2

De-arrangement

Number of arrangement of m things in a row so that none of them occupies its original place is m!{111!+12!13!+.+(1)m1 m!}

Exponent of prime p in n !

Ep(n!)=[np]+[np2]+[np3]+.+[npk] where k is the largest positive integer such that pkn<pk+1

Solved examples

1. The number of zeroes at the end of 100 ! is

(a) 23

(b) 24

(c) 25

(d) None of these

Show Answer

Solution:

[1005]+[10052]+[10053]+

=20+4+0

=24

Answer: (b)

2. The total number of integral solutions of the triplet (x,y,z) for the equation xyz=24 is

(a) 30

(b) 60

(c) 120

(d) None of these

Show Answer

Solution:

24.1.13!2!=312.2.13!=66.4.13!=68.3.13!=66.2.23!2!=34.3.23!=6 Total =30

30 positive integral solutions

Total number of integral solutions with negative integers included is 30×4=120

Answer: (c)

3. The total number of squares in a chess board is

(a) 64

(b) 65

(c) 204

(d) None of these

Show Answer

Solution:

12+22+32++82=8(8+1)(16+1)6=204

Answer: (c)

4. 20 lines pass through a given plane. The maximum number of parts in which the plane can be divided is

(a) 210

(b) 211

(c) 212

(d) None of these

Show Answer

Solution:

 Use 1+n=1+20=1+20.212=211

Answer: (b)

5. The number of quadrilaterals that can be formed using 10 points in a plane out of which 4 are collinear is

(a) 210

(b) 209

(c) 185

(d) None of these

Show Answer

Solution:

10C44C44C36C1=185

Answer: (c)

6. The total number of distinct rational numbers x such that 0<x<1 and x=pq where p,q{1,2,3,4,5,6} is

(a) 15

(b) 13

(c) 11

(d) None of these

Show Answer

Solution:

Values of p Possible rational numbers
1 12,13,14,15,16
2 23,24,25,26
3 34,35,36
4 45,46
5 56

Out of 15 possible rational numbers, only 11 are distinct.

Answer: (c)

7. The sum of 5 digit number in which only odd digits occur without repetition is

(a) 277775

(b) 555550

(c) 1111100

(d) None of these

Show Answer

Solution:

Sum of n digit numbers

=( Sum of digits )(10n1)101(n1)!=(1+3+5+7+9)(1051)101(51)!=25×11111×24=6666600

Answer: (d)

Practice questions

1. An n digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2,5 and 7. The smallest value of n for which this is possible is

(a) 6

(b) 7

(c) 8

(d) 9

Show Answer Answer: (b)

2. Match the following:

Consider all possible permutations of the letters of the word ENDEANOEL

Column I Column II
(a) The number of permutations containing the word ENDEA is (p) 5!
(b) The number of permutations in which the letteE occurs in the first and last position is (q) 2×5!
(c) The number of permutations in which none of the lette D,L,N occurs in the last five positions is (r) 7×5!
(d) The number of permutations in which the letter A,E,O occur only in odd positions is (S) 21×5!
Show Answer Answer: a\rarr p; b \rarr s; c \rarr q; d \rarr q

3. Five balls of different colors are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways we can place the balls so that no box is empty, is

(a) 116

(b) 126

(c) 144

(d) 150

Show Answer Answer: (d)

4. A student is allowed to select atmost n books from a collection of (2n+1) books. If number of ways in which he can select atleast one book is 63 , then n=

(a) 3

(b) 4

(c) 6

(d) 5

Show Answer Answer: (a)

5. A rectangle with sides 2 m1,2n1 is divided into squares of unit length by drawing lines parallel to sides of a rectangle. The number of rectangles with odd side length is

(a) (m+n+1)2

(b) mn(m+1)(n+1)

(c) m2n2

(d) 4m+n1

Show Answer Answer: (c)

6. Out of 5 apples, 10 mangoes and 15 oranges, the number of ways of distributing 15 fruits each to two persons, is

(a) 56

(b) 64

(c) 66

(d) 72

Show Answer Answer: (c)

7. Match the following

Column I Column II
(a). The number of positive integral solutions of the equati x1x2X3x4x5=1050 is λ, then λ is divisible by (p). 3
(b). Let be the element of the set A={1,2,3,5,6,10,15,30} anx1,x2,x3 be integers such that x1x2x3=y.If λ be the numbeof integral solutions of x1x2x3=y, then λ is divisible by (q). 4
(c). Let a be a factor of 120 . If λ be the number of positive integral solutions ox1x2x3=a, then λ is divisible by (r). 5
(s). 8
(t). 16
Show Answer Answer: a \rarr p, r; b \rarr q, s, t; c \rarr q, s, r, t

8. The maximum number of points into which 4 circles & 4 straight lines intersect is

(a) 26

(b) 50

(c) 56

(d) 72

Show Answer Answer: (b)

9. A is a set containing n elements. A subset P1 is chosen and A is reconstructed by replacing the elements of P1. The same process is repeated for subsets P2,.Pm with m>1. The number of ways of choosing P1,P2,.,Pm, so that P1P2Pm=A is

(a) (2m1)mn

(b) (2n1)m

(c) m+nCm

(d) None of these

Show Answer Answer: (d)

10. Number of points having position vector ai^+bj^+ck^ where a,b,c{1,2,3,4,5,} such that 2a+3b+5c is divisible by 4 is

(a) 70

(b) 140

(c) 210

(d) 280

Show Answer Answer: (a)

11. Read the passage and answer the following questions.

A is a set containing n elements. A subset P of A is chosen and the set A is reconstructed by replacing the elements of P. A subset Q of A is chosen again. Find the number of ways of choosing P & Q when

(i) Q is subset of P is

(a) 3n

(b) 2n

(c) n.3n1

(d) None of these

Show Answer Answer: (a)

(ii) P&Q contain just one element is

(a) 2n

(b) 3n

(c) n3n1

(d) None of these

Show Answer Answer: (c)

(iii) P=Q is

(a) 2n

(b) 3n

(c) n.3n1

(d) None of these

Show Answer Answer: (a)