IIT JEE

The number of solutions of x_{1}+x_{2}+x_{3}
= 51 (x_{1},x_{2},x_{3} being odd natural numbers) is

(A) 300

(B) 325

(C) 330

(D) 350

Let odd natural numbers be 2a – 1, 2b-1, 2c-1

where a, b, c are natural numbers

2a – 1+ 2b-1+2c-1=51

&⇒ a
+ b + c = 27

a ≥ 1, b ≥ 1, c ≥ 1 ….(1).

No. of solutions of (1) is coefficient of x^{24}
in (1-x)^{-3}.

= ^{26}C_{2} = 13 x 25 = 325

A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen. The number of ways of chosen P and Q so that P ÇQ = f is.

(A) 2^{2n}
– ^{2n}C_{n}

(B) 2^{n}

(C) 2^{n} –
1

(D) 3^{n}

Let A = { a_{1}, a_{2}, a_{3},
. . . , a_{n}} . For a_{i} ∈
A, we have the following choices:.

(i) a_{i} ∈ P and a_{i}∈Q (ii) a_{i} ∈ P and a_{i}∈Q

(iii) a_{i} ∈P and a_{i}∈Q (iv) a_{i} ∈ P and a_{i}∈Q

Out of these only
(ii) , (iii) and (iv) imply a_{i} ∈
P Ç Q. Therefore, the
number of ways in which none of a_{1}, a_{2}, . . .a_{n}
belong to P Ç Q is 3^{n}
. .

Let p be a prime number such that p ≥ 3. Let n = p! + 1. The number of primes in the list n+1, n+2, n+ 3, . . . . n + p –1 is .

(A) p –1

(B) 2

(C) 1

(D) none of these

For 1 £ k £ p –1, n+ k = p! +k +1, is clearly divisible by k +1.

Therefore, there is no prime number in the given list.

The number of ways in which a mixed double game can be arranged amongst 9 married couples if no husband and wife play in the same game is

(A) 756

(B) 1512

(C) 30 24

(D) none of these.

We can choose two men out of 9 in ^{9}C_{2}
ways. Since no husband and wife are to play in the same game, two women out of
the remaining 7 can be chosen in ^{7}C_{2} ways. If M_{1},
M_{2}, W_{1} and W_{2} are chosen, then a team may
consist of M_{1 }and W_{1} or M_{1} and W_{2}.
Thus the number of ways of arranging the game is.

(^{9}C_{2})
(^{7}C_{2})(2) = =
1512.

Number of even divisions of 504 is

(A) 12

(B) 24

(C) 6

(D) 18

504 = 2^{3} ´ 3^{2} ´
7.. Any even divisor of 504 is of the form 2^{i} ´ 3^{i}
´ 7^{k},
where I £ i£ 3,
0 £ j £ 2,
0 £ k£ 1.
Thus total number of even divisors is 3 ´3´ 2
= 18.

Ten persons are arranged in a row. The number of ways of selecting four persons so that no two persons sitting next to each other are selected is.

(A) 34

(B) 36

(C) 35

(D) none of these

To each selection of 4 persons we associated a binary sequence of the form 1001001010 where 1(0) at ith place means the ith person is selected (not selected).

There exists one-to-one correspondence between the set of selections of 4 persons and set of binary sequence containing 6 zeros and 4 ones.

We are interested in the binary sequences in which 2 ones are not consecutive. We first arrange 6 zeros.

0 0 0 0 0 0

This can be done in just one way. Now, 4 ones can be arranged at any of the 4 places marked with a cross in the following arrangement.

´ 0 ´ 0 ´ 0 ´ 0 ´ 0 ´ 0.

We can arrange 4 1’s at 7
places in ^{7}C_{4} = 35 ways.

A set contains (2n + 1) elements. The number of subset of the set which contain at most n elements is .

(A) 2^{n}

(B) 2^{n + 1}

(C) 2^{n –1 }

(D) 2^{2n}

The number of subsets of the set which contain at most n elements is

(say)

we have

No. of different garlands using 3 flowers of one kind and 12 flowers of second kind is .

(A) 19

(B) 11! ´ 2!

(C) ^{14}C_{2 }

(D) none of these

Number of different garlands will be equal to number of different solutions of the equation a + b + c = 12 without taking order of a, b and c into consideration. .

The number of ways in which letters of the word ARGUMENT can be arranged so that both the corners are filled by consonants only is

(A) 3! ´ 5!

(B) 14400

(C) 41000

(D) None of these

Vowels of the word ARGUMENT divide the consonants in
four parts, whose numbers are say y_{1}, y_{2}, y_{3}
and y_{4} where y_{1} + y_{2} + y_{3} + y_{4}
= 5,

y_{1} ≥ 1, y_{4} ≥1, y_{2} ≥ 0,
y_{3} ≥0

Number of solutions of this equation is 20. Hence required number is 20 ´ 5!´ 3! = 864000 (as consonants can be arranged in 5! ways and vowels can be arranged in 3! Ways).

The number of signals that can be generated by using 6 differently coloured flags, when any number of them may be hoisted at a time is

(A) 1956

(B) 1957

(C) 1958

(D) 1959

When one flag is used, the number of signals
that can be generated is ^{6}P_{1}. When two flags are used,
the number of signals that can be generated is ^{6}P_{2}. when
three flags are used, the number of signals that can be generated is ^{6}P_{3}
and so on. Hence, the number of different signals that can be generated is.

^{6}P_{1}
+ ^{6}P_{2} +^{ 6}P_{3} +^{ 6}P_{4}
+^{ 6}P_{5} +^{ 6}P_{6 }= 6 + 30 + 120 + 360 +
720 + 720 = 1956