**1) If x ∈ N and 3 x+1C _{2}+P_{2 }x = 4 xP_{2} 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 × 10^{4} + 10 × 18 × 10^{3} + 10 × 18 × 10^{2} + 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)**∑ 9P _{i }**(b)

**∑ 9P**(c)

_{i }**2**(d)

^{9}– 1**2**

^{10}– 1** i=1 i=1**

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

x_{1} , x_{2} ,…. x_{k }(1 ≤ K ≤ 9 and 9 ≥ x_{1} ≥ x_{2} > …. > x_{k} ≥ 1)

Total = 9C_{1} + 9C_{2} + ….. + 9C_{9} = 2^{9} – 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) **12 ^{4}** (b)

**28×12**(c)

^{3}**16×12**(d)

^{3}**4×12**

^{3}(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 6×12^{3} .

Case-2 : In Each one of the four pairs the left and right shoes have opposite colours. This can occur is 12^{4} 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) **2 ^{3}×3^{2} ** (b)

**2**(c)

^{3}×3^{3}**2**(d)

^{3}×3^{4}**2**

^{4}×3^{3}(d) The first 3 digits can be filled in 6^{3} 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 = 2×6^{3} = 2^{4}×3^{3}

**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) **2 ^{50}(2^{50}-1)** (b)

**2**(c)

^{100}-1**2**(d)

^{50}-1**2**

^{100}-2^{75}(a) Total number of non-empty subsets of S is 2^{100} – 1 .

Total no.of non empty subsets of S in which each elements is odd is 2^{50} – 1

Required number = (2^{100} – 1) – (2^{50} – 1)

**8) Let A be the set of 4-digit numbers a _{1}a_{2}a_{3}a_{4} where a_{1} > a_{2} > a_{3} >a_{4} , 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 = 10C_{4} = 210.

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

(a)** 41 (9 ^{3})** (b)

**37 (9**(c)

^{3})**7 (9**(d)

^{4})**41 (9**

^{4})(a) 5 digit numbers having 7 in 1^{st} place = 9^{4}

In 2^{nd}, 3^{rd}, 4^{th}, 5^{th} places is 4 x 8 x 9^{3}

Total number of 5 digit numbers having 7 exactly once is = 41 (9^{3}).

**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)]

= 4C_{0} × 4C_{4} + 4C_{1} ×4C_{3} + 4C_{2} × 4C_{2}

= 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) **2 ^{2n}** (b)

**2**

^{2n+1 }(c)

**2**(d)

^{2n-1}**2(2**

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

This can be done in 2n+1C_{n-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 6C_{2} × 4C_{2} ways

The numbers of arrangements is 3! × 6C_{2} × 4C_{2} × (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,M_{1},M_{2},M_{3}, , , ,

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

When twins are not consecutive, we get 3 arrangements when M_{i} ‘s are in cycle M_{1}M_{2}M_{3 }and

similarly, also for the cycle, M_{1}M_{3}M_{2}

Total = 12 arrangements.

**14) Let A = {x _{1} , x_{2} , x_{3} ,…,x_{7}} , B = {y_{1}y_{2}y_{3}}. The total number of functions f :A→B that are onto and there are exactly three elements x in A such that f(x) = y_{2}, is equal to**

(a) **14.7C _{2}** (b)

**14.7C**

_{3}^{ }(c)

**7.7C**(d)

_{2}**7.7C**

_{3}(b) A = { x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7} }, B = { y_{1},y_{2},y_{3} }

f :A →B is onto ∋ f(x) = y_{2}

Exactly 3 elements x in is y_{2} .

This can be done in 7C_{3} ways

Remain A four elements in B 2 elements

∴ 2^{4} – 2C_{1} (2 – 1)^{4} = 14

Total no. of onto functions 7C_{3} = 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) = 2007^{2} = 3^{4}×223^{2}

The number of pairs is equal to the number of divisors of 2007^{2 }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 x _{1}, x_{2}, x_{3}, x_{4}, x_{5} = 2310 are**

(a) **5 ^{5} ** (b)

**6.5**(c)

^{5 }**16.5**(d)

^{5}**5**

^{6}(a) x_{1}, x_{2}, x_{3}, x_{4}, x_{5} = 2310 3′ 7′ 10′ 11′ each of 3, 7, 10, 11 can be distributed at 5 places in 5 ways + ve integral sols are 5^{5}

(i) Two are negative and 3 positive then 5C_{3 }5^{5} ways

(ii) Four are negative and 1 positive then 5C_{4 }5^{5 }ways

**18) Total number of positive integral solutions of 15! < x _{1} + x_{2} + x_{3} ≤ 20, is equal to **

(a)** 1125** (b) **1150 ^{ }** (c)

**1245**(d)

**685**

(d) 15 < x_{1} + x_{2} + x_{3} ≤ 20

⇒ x_{1} + x + x_{3} = 16 + r

r = 0, 1, 2, 3, 4

Number of positive integral solutions of

x_{1} + x_{2} + x_{3} = 16 + r is 13 + r + 3 – 1C_{13 + r , }15 + rC_{13 + r} = 15 + rC_{2}

4

Required number of solutions = ∑ 15+rC_{2} = 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 3^{2}, 7^{2}, 3^{2}. 5^{3} and 3^{1}. 2^{7}.

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 = 2 ^{P1 }3^{P2 }5^{P3 }7^{P4 }11^{P5 }13^{P6} then ∑ P_{r} is**

** ****r=1**

(a) **24** (b) **23**^{ } (c) **22** (d) **21**

(a) 15! = 2^{11}×3^{6}×5^{3}×7^{2}×11^{1}×13^{1}

6

∴ ∑ P_{r} = 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 3C_{1} ways, there are 3C_{1} (9 × 9) = 3 × 9^{2 }such numbers.

Next, 3 can occur in exactly two places in (3C_{2})(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 ×9^{2}) + 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

= 10^{5} × 5s 500000/(60×60×13) days = 10.7days

Hence, 11 days are enough to open the safe.