Permutations and Combinations — Page 10 | JEE Mathematics

Q91

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is :

A 500

B 350

C 200

D 300

Correct Answer

Option D

Solution

From 5 girls 2 girls can be selected = 5C2 ways From 7 boys 3 boys can be selected = 7C3 way $\therefore$ Total number of ways we can select 2 girls and 3 boys = 5C2 $\times$ 7C3 ways When two boys A and B are chosen in a team then one more boy will be chosen from remaining 5 boys.

So, no of ways 3 boys can be chosen when A and B should must be chosen = 5C1 ways $\therefore$ Total number of ways a team of 2 girl and 3 boys can be made where boy A and B must be in the team = 5C1 $\times$ 5C2 ways $\therefore$ Required number of ways = Total number of ways $-$ when A and B are always included.

= 5C2 $\times$ 7C3 $-$ 5C1 $\times$ 5C2 = 300

Q92

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

A 145

B 165

C 155

D 135

Correct Answer

Option C

Solution

1 Captain, 1 vice-captain are already present $\Rightarrow$ We need to select 8 players such that atleast 3 batsman and bowler must be there .tg .tg Batsman Bowler Number of ways 3 5 ${ }^6 C_3 \cdot{ }^5 C_5=20$ 4 4 ${ }^6 C_4 \cdot{ }^5 C_4=75$ 5 3 ${ }^6 C_5 \cdot{ }^5 C_3=60$ Total = 155 ways

Q93

The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus

A \cup B \cup C = S,\,A \cap B = B \cap C = A \cap C = \phi

. The number of ways to partition S is

A

{{12!} \over {{{(4!)}^3}}}\,\,

B

{{12!} \over {{{(4!)}^4}}}\,\,

C

{{12!} \over {3!\,\,{{(4!)}^3}}}

D

{{12!} \over {3!\,\,{{(4!)}^4}}}

Correct Answer

Option A

Solution

The total number of ways is

{}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}

Q94

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

A 5040

B 6210

C 385

D 1110

Correct Answer

Option C

Solution

A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.

Case 1 : When he give vote to only 1 candidate then no ways =

{}^{10}{C_1}

Case 2 : When he give vote to 2 candidates then no ways =

{}^{10}{C_2}

Case 3 : When he give vote to 3 candidates then no ways =

{}^{10}{C_3}

Case 4 : When he give vote to 4 candidates then no ways =

{}^{10}{C_4}

So, total no of ways he can give votes =

{}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}

= 385 Note : Here we use addition rule as he can vote any one of those four rules.

Whenever there is "or" choices, we use addition rule.

Q95

The sum of integers from 1 to 100 that are divisible by 2 or 5 is :

A 3000

B 3050

C 3600

D 3250

Correct Answer

Option B

Solution

According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2 $\times$ 5 = 10.

Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100 This is an A.P where first term = 2, last term = 100 and total terms = 50.

$\therefore$ Sum of the numbers divisible by 2 =

{{50} \over 2}\left[ {2 + 100} \right]

= 25 $\times$ 102 = 2550 Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100 $\therefore$ Sum of the numbers divisible by 5 =

{{20} \over 2}\left[ {5 + 100} \right]

= 10 $\times$ 105 = 1050 And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100 $\therefore$ Sum of the numbers divisible by 10 =

{{10} \over 2}\left[ {10 + 100} \right]

= 5 $\times$ 110 = 550 $\therefore$ Required sum = 2550 + 1050 - 550 = 3050

Q96

If

{{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}

= 11, then n satisfies the equation :

A n2 + 3n − 108 = 0

B n2 + 5n − 84 = 0

C n2 + 2n − 80 = 0

D n2 + n − 110 = 0

Correct Answer

Option A

Solution

{{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11

$\Rightarrow$

{{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}

$\Rightarrow$ (n + 2)!

= 11.6!

(n $-$ 2)!

$\Rightarrow$ (n + 2) (n + 1) n (n $-$ 1) = 11.6!

$\Rightarrow$ (n + 2) (n + 1) n (n $-$ 1) = 11 .

6 .

5 .

4 .

3 .

2 . 1 $\Rightarrow$ (n + 2) (n + 1) n (n $-$ 1) = 11 .

10 .

9 . 8 $\therefore$ n = 9 This value of n satisfy the equation, n2 + 3n $-$ 108 = 0

Q97

If the four letter words (need not be meaningful ) are to be formed using the letters from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is :

A

{{11!} \over {{{\left( {2!} \right)}^3}}}

B 110

C 56

D 59

Correct Answer

Option D

Solution

Here total no of different letters present are, (1) One M (2) Three E (E E E) (3) One D (4) One I (5) One T (6) Two R (R R) (7) Two A (A A) (8) Two N (N N) In the four letter word first letter is R and last letter is E.

$\therefore$ Word is = R _ _ E Now remaining letters are, M, EE, D, I, T, R, AA, NN Those 2 empty places can be filled with identical letters [EE, AA, NN] in 3 ways.

Or two empty places can be filled with distinct letters [M, E, D, I, T, R, A, N] in

{}^8{C_2} \times 2!

ways. $\therefore$ Total no of words = 3 +

{}^8{C_2} \times 2!

= 59

Q98

The sum

\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)}

is equal to :

A (11)!

B 10

\times

(11!)

C 101

\times

(10!)

D 11

\times

(11!)

Correct Answer

Option B

Solution

\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!

= \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!}

= \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]}

= \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} }

= \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!}

= \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]

- \sum\limits_{r = 1}^{10} {r\,.\,r!}

$=$ 11.11! $-$ 1.1! $-$

\sum\limits_{r = 1}^{10} {r\,.\,r!}

$=$ (11.11! $-$ 1) $-$

\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!

$=$ (11.11! $-$ 1) $-$

\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]}

$=$ (11.11! $-$ 1) $-$

\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)! - r!} \right]}

$=$ (11.11!

$-$ 1) $-$ [(2!

$-$ 1!)

+ (3!

$-$ 2!)

+ . . . .

+ (11!

$-$ 10!)]

$=$ (11.11!

$-$ 1) $-$ (11!

$-$ 1) $=$ 11.11!

$-$ 11!

$=$ 11!

(11 $-$ 1) $=$

10\,.\,\left( {11!} \right)

Q99

The value of

\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)

is equal to :

A 560

B 680

C 1240

D 1085

Correct Answer

Option B

Solution

We know,

{{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}

$\therefore$

{{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}

$\therefore$

\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)

$=$

\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)

= \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)}

= 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} }

= 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}

$=$ 120 $\times$ 16 $-$ 40 $\times$ 31 $=$ 680

Q100

Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is :

A 9

B 18

C 36

D 32

Correct Answer

Option C

Solution

Area =

{1 \over 2}

h. k = 50 h. k = 100 h. k = 22 . 52 Total divisors = (2 + 1) (2 + 1) = 9 if h > 0, k > 0 But

\begin{array}{ll}{h > 0,} & {k < 0} \\ {h < 0,} & {k > 0} \\ {h < 0,} & {k < 0} \end{array}

all are possible so that total no. of positive case 9 + 9 + 9 + 9 = 36