Mathametics 09 – Permutation and combination

1) If x ∈ N and 3 x+1C2+P2 x = 4 xP2 then

(a) x=5               (b) x=3              (c) x = 3n+1               (d) x=10

(b) 3(x+1)! ⁄ 2!(x+1)! + 2!× x = 4 x! ⁄ (x-2)!

3 ⁄ 2(x+1) +2x = 4x(x+1)

⇒(3 ⁄ 2)x + (3 ⁄ 2) + 2 – 4x + 4 =0

⇒(-5x ⁄ 2)+(15 ⁄ 2) = 0

x = 3 

2) Sum of all the five digits natural numbers formed using all the digits {0, 1, 2, 3, 4}

(a) 2599980               (b) 2666640              (c) 1999980              (d) none of these

(a) Sum of all the number is 10 × 24 × 104 + 10 × 18 × 103 + 10 × 18 × 102 + 10 × 18 × 10 + 10 × 18 × 1

= 2599980

3) The no. of rational numbers lying in the interval (2002, 2003) all whose digits 

after the decimal point are non-zero and are in decreasing order

      9                              9

(a) 9Pi                       (b) 9Pi                (c) 29 – 1               (d) 210 – 1

     i=1                            i=1

(c) A rational number of the desired category is of the form 2002.

x1 , x2 ,…. xk (1 ≤ K ≤ 9 and 9 ≥ x1 ≥ x2 > …. > xk ≥ 1)

Total = 9C1 + 9C2 + ….. + 9C9 = 29 – 1

4) There are four pairs of shoes of different sizes. Each of the 8 shoes can be coloured with one of the four colours : Black, Brown, White and Red. The number of ways the shoes can be coloured so that in atleast three pairs, the left shoe and the right shoe do not have the same colour is

(a) 124               (b) 28×123              (c) 16×123               (d) 4×123

(b) We discuss two disjoint possibilities

Case -1 : In Exactly three pairs the shoes of a pair have opposite colours. The number of ways this occur is 6×123 .

Case-2 : In Each one of the four pairs the left and right shoes have opposite colours. This can occur is 124 ways.

5) The letters of the word “DRAWER” are arranged in alphabetical order. The number of arrangements that precede the word “ REWARD” is

(a) 241               (b) 242              (c) 247               (d) 248

(c) The letters of the word ‘DRAWER’ in alphabetical order are A, D, E, R, R and W.

The number of words that precede the word “REWARD” is

3×(5!/2!) + 2×4! +3(3!) +1 = 247

6) The number of four digit numbers that can be formed using the digits 1,2,3,4,5,6 that are divisible by 3,when repetition of digits is allowed, is

(a) 23×32               (b) 23×33              (c) 23×34               (d) 24×33

(d) The first 3 digits can be filled in 63 ways. After filling the first 3 digits, the last digit can be filled using

1, 2, 3, 4, 5, 6 in 6 ways.

But the six numbers so formed are consecutive integers out of which only two are divisible by 3.

Hence the number of choices = 2×63 = 24×33

7) Let S = {1,2,3,…100} . The number of non-empty subsets A of S such that the product of elements in A is even is

(a) 250(250-1)               (b) 2100-1              (c) 250-1              (d) 2100-275

(a) Total number of non-empty subsets of S is 2100 – 1 .

Total no.of non empty subsets of S in which each elements is odd is 250 – 1

Required number = (2100 – 1) – (250 – 1)

8) Let A be the set of 4-digit numbers a1a2a3a4 where a1 > a2 > a3 >a4 , then n(A) is equal to

(a) 126               (b) 84              (c) 210              (d) none of these

(c) Any selection of four digits from the ten digits 0, 1, 2, 3, … 9 gives one such number.

SO, the required number of numbers = 10C4 = 210.

9) The number of 5 digit numbers that contain 7 exactly once is

(a) 41 (93)               (b) 37 (93)               (c) 7 (94)               (d) 41 (94)

(a) 5 digit numbers having 7 in 1st place = 94

In 2nd, 3rd, 4th, 5th places is 4 x 8 x 93

Total number of 5 digit numbers having 7 exactly once is = 41 (93).

10) 4 points out of 8 points in a plane are collinear. Number of different quadrilateral that can be formed by joining them is

(a) 56              (b) 53               (c) 76              (d) 60

(b) Number of ways = [collinear (o) and non collinear (40)]

OR [collinear (1) and non collinear (3)]

OR [collinear (2) and non collinear (2)] 

= 4C0 × 4C4 + 4C1 ×4C3 + 4C2 × 4C2 

= 1+16+36 =53

11) The number of ways of choosing n objects out of 3n+1 objects of which ‘n’ are identical and (2n+1) are distinct is

(a) 22n              (b) 22n+1               (c) 22n-1              (d) 2(2n)

(a) If we choosevk(0 ≤ k ≤ n) identical objects, then we must choose n-k distiact objects.

This can be done in 2n+1Cn-k ways.

Thus, the required number of ways.

12) Four men and six women are to be seated along a round table. The number of cases where every two men are separated by two women is

(a) 3420             (b) 4320               (c) 1710              (d) None of these

(b) 4 men may be seated in 3! Ways.

The pairs of women can be selected in 6C2 × 4C2 ways

The numbers of arrangements is 3! × 6C2 × 4C2 × (2!)

6×15×6×8 = 4320 ways

13) Five boys, of whom two are identical twins, are to be seated around a round table. The number of ways is

(a) 60             (b) 12               (c) 81              (d) None of these

(b) T,T,M1,M2,M3, , , ,

With twins together we get(4 – 1)!  = 6 arrangements.

When twins are not consecutive, we get 3 arrangements when Mi ‘s are in cycle M1M2M3 and

similarly, also for the cycle, M1M3M2

Total = 12 arrangements.

14) Let A = {x1 , x2 , x3 ,…,x7} , B = {y1y2y3}. The total number of functions f :A→B that are onto and there are exactly three elements x in A such that f(x) = y2, is equal to

(a) 14.7C2             (b) 14.7C3               (c) 7.7C2              (d) 7.7C3

(b) A = { x1,x2,x3,x4,x5,x6,x7 }, B = { y1,y2,y3 }

f :A →B is onto ∋ f(x) = y2

Exactly 3 elements x in is y2 .

This can be done in 7C3 ways

Remain A four elements in B 2 elements

∴ 24 – 2C1 (2 – 1)4 = 14

Total no. of onto functions 7C3 = 14

15) A round table conference is to be held between 20 delegates of 2 countries. In how many ways can they be seated if two particular delegates are always to sit together?

(a) 2 (18!)             (b) 19! – 2(18!)               (c) 19!               (d) 18!

(a) Since two delegates always sit together. Hence 20 – 2 = 18 and

Total 19 delegates can be sited in round table.

∴Number of ways = 2.(18)!

16) The number of the positive integer pairs (x,y) such that (1/x)+(1/y) = 1/2007 where x < y is

(a) 5             (b) 6               (c) 7              (d) 8

(c) (1/x)=(1/y) = (1/2007)

⇒(x+y)2007 = xy

⇒xy – 2007x – 2007y = 0

(x – 2007)(y – 2007) = 20072 = 34×2232

The number of pairs is equal to the number of divisors of 20072 that is (4+1)×(2+1) = 15

Since,x < y,

so required number of pairs = 7

17) The number of integral solutions of the equation x1, x2, x3, x4, x5 = 2310 are

(a) 55             (b) 6.5              (c) 16.55              (d) 56

(a) x1, x2, x3, x4, x5 = 2310 3′ 7′ 10′ 11′ each of 3, 7, 10, 11 can be distributed at 5 places in 5 ways + ve integral sols are 55

(i) Two are negative and 3 positive then 5C3 55 ways

(ii) Four are negative and 1 positive then 5C4 55 ways

18) Total number of positive integral solutions of 15! < x1 + x2 + x3 ≤ 20, is equal to

(a) 1125             (b) 1150               (c) 1245              (d) 685

(d) 15 < x1 + x2 + x3 ≤ 20

 

⇒ x1 + x + x3 = 16 + r

 

r = 0, 1, 2, 3, 4

 

Number of positive integral solutions of

 

x1 + x2 + x3 = 16 + r is 13 + r + 3 – 1C13 + r , 15 + rC13 + r = 15 + rC2

                                                      4

Required number of solutions = 15+rC2 = 685

                                                    r=0

19) The number of divisors of 441, 1125 and 384 are in

(a) A.P             (b) G.P               (c) H.P.               (d) None of these

(b) The given numbers can be written as 9 × 49,9 × 125, 3 × 128 or 32, 72, 32. 53 and 31. 27.

Hence the number of divisors are

3.3, 3.4, 2.8 or 9, 12, 16

Which are in G.P. of r = 4/3.

                                                                 6

20) If 15 = 2P1 3P2 5P3 7P4 11P5 13P6 then   Pr is

                                                               r=1


(a) 24             (b) 23               (c) 22               (d) 21

(a) 15! =  211×36×53×72×111×131

    6

Pr = 11+6+3+2+1+1 = 24

   r=1

21) The number of times the digit 3 will be written when listing the integers from 1 to 1000, is

(a) 269             (b) 300               (c) 271              (d) 302

(b) Since 3 does not occur in 1000, we have to count the number of times 3 occurs when we list the integers from 1 to 999.

Any number between 1 and 999 is of the form xyz where 0 ≤ x,y,z ≤ 9

Let us first count the number in which 3 occurs exactly once. Since 3 can occur at one place in 3C1 ways, there are 3C1 (9 × 9) = 3 × 92 such numbers.

Next, 3 can occur in exactly two places in (3C2)(9) = 3 × 9 such numbers.

Lastly, 3 can occur in all three digits in one number only.

Hence, the number of times 3 occurs is 1 × (3 ×92) + 2 × (3 × 9) + 3 × 1 = 300

22) The lock of a safe consists of five discs each of which features the digits 0, 1, 2, ….., 9. The safe can be opened by dialing a special combination of the digits. The number of days sufficient enough to open the safe, if the work day lasts 13 h and 5 s are needed to dial one combination of digits is

(a) 9             (b) 10               (c) 11              (d) 12

(c) Total time required = (total number of dials required to sure open the lock) × 5s

= 105 × 5s 500000/(60×60×13) days = 10.7days

Hence, 11 days are enough to open the safe.

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