PERMUTATION AND COMBINATIONS - 1 (Properties of nPr and nCr)

Fundametal principle of counting

(i) Addition principle : If an operation can be performed in ’ m ’ different ways and another operation which is independent of the first operation, can be performed in ’ n ’ different ways then either of them can be performed in (m+n) ways.

(ii) Multiplication Principle: If an operation can be performed is ’ m ’ different ways; following with a second operation can be performed in ’ n ’ different ways, then the two operations in succession can be performed in ‘mn’ ways.

Factorial

The continued product of first n natural numbers is n!.

n!=1.2.3 n

value of 0 ! is 1

Exponent of prime p is n !

Let n be a positive integer and p, a prime number. Then the exponent of p is n ! is given by Ep(n!)=[np]+[np2]+[np3]++[nps] where s in the largest positive integer such that ps<n<ps+1

Permutation

Number of permutations of n distinct things taking r(1rn) at a time is denoted by nPr.

nPr=n!(nr)!

  • Number of ways of filling r places using n things if repetition is allowed =nr

Circular Permutation

Number of circular permutations of n things =(n1) !

Number of circular permutations of n different things taken r at a time =nPrr

Number of circular permutations of n different things when clockwise and anticlockwise circular permutations are considered as same is (n1)!2

Note: When position are marked, circular arrangement is assumed to be linear.

Combination

Number of combinations of n distinct things taking r at a time is denoted by nCr.

nCr=n!r!(nr)!

Note: nPr=r ! (nCr)

Properties of nPr and nCr

i. nPn=n!;nCo=nCn=1

ii. nPr=n1Pr1+rn1Pr1

iii. r=1nrrPr=n+1Pn+11

iv. nCr=nCs either r=s or r+s=n

v. nCr is greatest, {r=n2; if n is even r=n±12; if n is od(d). 

vi. nCr=nCnr=nrn1Cr1

vii. nCrnCr1=nr+1r

viii. nCr1+nCr=n+1Cr

Solved examples

1. How many 5-digit numbers divisible by 3 can be formed using digits 0,2,4,6,8, 9, if repetition not allowed?

Show Answer

Solution:

Sum of digits =0+2+4+6+8+9=29

so 5 -digit numbrs can be formed using the digits 0,4,6,8,9 or 0,2,4,5,9 as sum of digits is divisible by 3 .

total number of numbers formed is

2(4.4.!)=192

Answer: 192

2. How many 5 -digit numbers divisible by 4 can be formed using digits 0,2,4,7,8,9, if repetition not allowed?

Show Answer

Solution:

Number is divisible by 4 , if the last two digits of the number are a multiple of 4 , so we can have 04 , 08,20,24,28,40,48,72,80,84,92 at the last two places. Hence total number of numbers can be 5.(4P3)+6.3.3P2 as the first place cannot have zero.

the answer is 228 .

Answer: 228

3. How many 5 -digit numbers divisible by 6 can be formed using digits 0,2,4,5,6,8, if repetition not allowed.

Show Answer

Solution:

Sum of digits is

0+2+4+5+6+8=25

the numbers can be formed with the digits 0,2,5,6,8 as their sum is a multiple of 3 provided unit’s place has an even number. So it can be done in 4!+3.3.3!=78 ways.

4. The total number of odd natural numbers that can be formed with the digits 1,3,1,5,4,1,4 and are greater than 2 million are

(a) 120

(b) 160

(c) 180

(d) none

Show Answer

Solution:

Odd digits Even digits
1, 3, 5 4
1 4
1

Number of arrangements can be

Total no. =10+30+60+20+20+30+10=180

Answer: c

5. The total number of 4 digit numbers greater than 4000 , whose sum of digits is odd is

(a). 2800

(b). 3000

(c). 3600

(d). none of these

Show Answer

Solution:

Thousands place can be filled in 6 ways. Hundred’s and Ten’s place can be filled in 10 ways each. First 3 places given the sum either odd or even. In either case last place can be filled in 5 ways.

Number of 4 digit numbers is

=6×10×10×5=3000

Answer: b

6. Mohan writes a letter to five of his friends and addresses them. The number of ways in which the letters can be placed in the envelopes so that that three of them are in the wrong envelopes is

(a). 44

(b). 119

(c). 21

(d). 20

Show Answer

Solution:

5C23!(111!+12!13!)=20 (dearrangement of n pairs)

Answer: d

7. In how many ways the squares of the figure given below be filled up with letters of the word ‘ROHINI’, so that each row contains atleast one letter.

Show Answer

Solution:

(8C62)×6!2!=26×360=9360 ways 

as six squares can be slected in such a way that no row is empty is (8C62)=26 ways.

Answer: 9360

Practice questions

1. Number of polynomials of the form x3+ax2+bx+c that are divisible by x2+1, where a,b,cε{1, 2,3,..9,10}

(a). 30

(b). 1000

(c). 10

(d). none of these

Show Answer Answer: (c)

2. The number of ways in which we can select four numbers from 1 to 30 so as to exclude every selection of four consecutive number is

(a). 27378

(b). 27405

(c). 27399

(d). none of these

Show Answer Answer: (a)

3. The number of ordered pairs of integers (x,y) satisfying the equation x2+6x+y2=4 is

(a). 2

(b). 8

(c). 6

(d). none of these

Show Answer Answer: (b)

4. How many six digit numbers are there in which sum of the digits is divisible by 5

(a). 18,0000

(b). 54,0000

(c). 5×105

(d). none of these

Show Answer Answer: (a)

5. Ten IIT and 2 DCE students sit in a row. The number of ways in which exactly 3 IIT students sit between 2 DCE students is

(a). 10C3×2!×3!×8 !

(b). 10!2!3!×8 !

(c). 5!×2!×9!×8 !

(d). none of these

Show Answer Answer: (a)

6. Let there are n3 circles. The value of n for which the number of radical centres is equal to the number of radical axis is (assume that all radical axis and radical centres exist and are different)

(a). 7

(b). 6

(c). 5

(d). none of these

Show Answer Answer: (c)

7. Total number of integers ’ n ’ such that 2n2000 and HCF of n and 36 is one, is equal to

(a). 666

(b). 667

(c). 665

(d). none of these

Show Answer Answer: (a)

8. Match the following:-

Column I Column II
a. The number of five digit numbers having the product of digits 20 is p. 77
b. A man took 5 space plays out of an engine to clean them. The number ofways in which he can place atleast two plays in the engine from where they came out is q. 31
c. The number of integers between 1 and 1000 inclusive in which atleast two consecutive digits are equal is r. 50
d. Value of 1151ij9iij s. 181
Show Answer Answer: a - r, b - q, c - s, d - p

9. The number of ordered triplets (a,b,c) such that LCM(a,b)=1000,LCM(b,c)2000 and LCM(c,a)=2000 is ..

Show Answer Answer: 70

10. Read the paragraph and answer the questions that follow :-

Number of ways of distributing n different things into r different groups is rn when blank groups are taken into account and is rnrC1(r1)n+rC2(r2)2.+(1)r1rCr1 when blank groups are permitte(d).

i. 4 candidates are competing for two managerial posts. In how many ways can the candidates be selected?

(a). 42

(b). 4C2

(c). 24

(d). none of these

Show Answer Answer: (b)

ii. 8 different balls can be distributed among 3 children so that every child receives at least one ball is

(a). 38

(b). 8C3

(c). 83

(d). none of these

Show Answer Answer: (d)

iii. 5 letters can be posted into 3 letter boxes in

(a). 35 ways

(b). 53 ways

(c). 5C3 ways

(d). none of these

Show Answer Answer: (a)