Permutations and Combinations — Page 5 | JEE Mathematics

Q41

There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is :

A 2250

B 2255

C 3000

D 1500

Correct Answer

Option A

Solution

$\therefore$ Total number of selection of 5 questions =

\left( {{}^5{C_1}.{}^5{C_2}.{}^5{C_2}} \right)

$\times$ 3 +

\left( {{}^5{C_1}.{}^5{C_1}.{}^5{C_3}} \right)

$\times$ 3 = 5 $\times$ 10 $\times$ 10 $\times$ 3 + 5 $\times$ 5 $\times$ 10 $\times$ 3 = 2250

Q42

If the number of five digit numbers with distinct digits and 2 at the 10th place is 336 k, then k is equal to :

A 6

B 8

C 4

D 7

Correct Answer

Option B

Solution

Except 1 and 2 there are eight options on first place and only one option at fourth place possible which is digit '2'.

Remaining three places can be filled by any three digits out of 1, 3, 4, 5, 6, 7, 8 and 9.

$\therefore$ No. of five digits numbers = 8 $\times$ 8 $\times$ 7 $\times$ 1 $\times$ 6 = 336 $\times$ 8 Also 336k = 336 $\times$ 8 $\Rightarrow$ k = 8

Q43

Let n > 2 be an integer. Suppose that there are n Metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is :

A 201

B 199

C 101

D 200

Correct Answer

Option A

Solution

Number of blue lines = Number of sides = n Number of red lines = number of diagonals = nC2 – n According to question,

{}^n{C_2} - n = 99n

\Rightarrow {{n\left( {n - 1} \right)} \over 2} - n = 99n

\Rightarrow {{n - 1} \over 2} = 99

\Rightarrow n = 201

Q44

If a, b and c are the greatest value of 19Cp, 20Cq and 21Cr respectively, then :

A

{a \over {11}} = {b \over {22}} = {c \over {21}}

B

{a \over {10}} = {b \over {22}} = {c \over {21}}

C

{a \over {10}} = {b \over {11}} = {c \over {42}}

D

{a \over {11}} = {b \over {22}} = {c \over {42}}

Correct Answer

Option D

Solution

( 19Cp)max = 19C9 or 19C10 = a ( 20Cp)max = 20C10 = b ( 21Cr)max = 21C10 or 21C11 = c 1 =

{a \over {^{19}{C_{10}}}} = {b \over {^{20}{C_{10}}}} = {c \over {^{21}{C_{10}}}}

$\Rightarrow$

{a \over {11}} = {b \over {22}} = {c \over {42}}

Q45

The total number of positive integral solutions (x, y, z) such that xyz = 24 is :

A 36

B 24

C 45

D 30

Correct Answer

Option D

Solution

x.y.z = 24

x.y.z = {2^3}.\,{3^1}

Three 2 has to be distributed among x, y and z Each may receive none, one or two $\therefore$ Number of ways =

{}^{3 + 3 - 1}{C_{3 - 1}}

=

^5{C_2}

ways Similarly one 3 has to be distributed among x, y and z $\therefore$ Number of ways =

{}^{1 + 3 - 1}{C_{3 - 1}}

=

^3{C_2}

ways Total ways =

^5{C_2}\,.{\,^3}{C_2}

= 30

Q46

A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed, is :

A 1050

B 575

C 560

D 1625

Correct Answer

Option D

Solution

Given, Number of Indians = 6 Number of foreigners = 8 Committee of at least 2 Indians and double number of foreigners is to be formed.

Hence, the required cases are (2I, 4F) + (3I, 6F) + (4I, 8F) =

{}^6{C_2} \times {}^8{C_4} + {}^6{C_3} \times {}^8{C_6} + {}^6{C_4} \times {}^8{C_8}

= (15 $\times$ 70) + (20 $\times$ 28) + (15 $\times$ 1) = 1050 + 560 + 15 = 1625

Q47

A natural number has prime factorization given by n = 2x3y5z, where y and z are such that y + z = 5 and y

-

1 + z

-

1 =

{5 \over 6}

, y > z. Then the number of odd divisions of n, including 1, is :

A 11

B 6

C 12

D 6x

Correct Answer

Option C

Solution

y + z = 5 ....... (1)

{1 \over y} + {1 \over z} = {5 \over 6}

\Rightarrow {{y + z} \over {yz}} = {5 \over 6}

\Rightarrow {5 \over {yz}} = {5 \over 6}

$\Rightarrow$ yz = 6 Also, (y $-$ z)2 = (y + z)2 $-$ 4yz $\Rightarrow$ (y $-$ z)2 = (y + z)2 $-$ 4yz $\Rightarrow$ (y $-$ z)2 = 25 $-$ 4(6) = 1 $\Rightarrow$ y $-$ z = 1 ..... (2) from (1) and (2), y = 3 and z = 2 for calculating odd divisor of p = 2x . 3y . 5z x must be zero P = 20 .

33 .

52 $\Rightarrow$ Total possible cases = (3050 + 3150 + 3250 + 3350 + ....

+ 3352) $\therefore$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12

Q48

Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let

\alpha

be the number of triangles having these points from different sides as vertices and

\beta

be the number of quadrilaterals having these points from different sides as vertices. Then (

\beta

-

\alpha

) is equal to :

A 717

B 795

C 1890

D 1173

Correct Answer

Option A

Solution

\alpha = {}^6{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^7{C_1}

= 378 + 315 + 270 + 210 = 1173

\beta = {}^5{C_1}{}^6{C_1}{}^7{C_1}{}^9{C_1} = 1890

$\therefore$

\beta - \alpha = 1890 - 1173 = 717

Q49

Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to :

A 5

B 2

C 4

D 6

Correct Answer

Option C

Solution

Total matches between boys of both team =

{}^7{C_1} \times {}^4{C_1} = 28

Total matches between girls of both team =

{}^n{C_1}\,{}^6{C_1} = 6n

Now, 28 + 6n = 52 $\Rightarrow$ n = 4

Q50

If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :

A 240

B 360

C 333

D 364

Correct Answer

Option C

Solution

Total number of triangles =

{}^{14}{C_3} - {}^3{C_3} - {}^5{C_3} - {}^6{C_3}

= 364 – 31 = 333