Permutations and Combinations — Page 2 | JEE Mathematics

Q11

The number of ways of selecting two numbers

a

and

b, a \in\{2,4,6, \ldots ., 100\}

and

b \in\{1,3,5, \ldots . ., 99\}

such that 2 is the remainder when

a+b

is divided by 23 is :

A 186

B 54

C 108

D 268

Correct Answer

Option C

Solution

a+b=23\lambda+2

\lambda=0,1,2,

...., but $\lambda$ cannot be even as

a+b

is odd

\lambda=1

(a, b)\to12

pairs

\lambda=3

(a,b)\to35

pairs

\lambda=5

(a,b)\to42

pairs

\lambda=7

(a,b)\to19

pairs

\lambda=9

(a,b)\to0

pairs $\vdots$ Total

=12+35+42+19=108

Q12

Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is :

A 125

B 105

C 374

D 625

Correct Answer

Option C

Solution

There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed Case 1 : First digit is 1 = 1 _ _ _ Possible numbers starting with 1 = 1 $\times$ 5 $\times$ 5 $\times$ 5 = 125 But this includes 1000 also which does not satisfy the given condition of being greater than 1000.

Hence there will be 124 numbers having 1 in the first place.

Case 2 : First digit is 2 = 2 _ _ _ Possible numbers starting with 2 = 1 $\times$ 5 $\times$ 5 $\times$ 5 = 125 Case 3 : First digit is 3 = 3 _ _ _ Possible numbers starting with 3 = 1 $\times$ 5 $\times$ 5 $\times$ 5 = 125 Total possible numbers = 124 + 125 + 125 = 374

Q13

Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are :

A 312

B 3125

C 120

D 216

Correct Answer

Option D

Solution

Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.

Here given numbers are 0, 1, 2, 3, 4 and 5.

Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.

Set 1 : Set is = (1, 2, 3, 4, 5).

Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3) So total no of arrangement = 1 $\times$ 2 $\times$ 3 $\times$ 4 $\times$ 5 = 5!

Set 2 : Set is = (0, 1, 2, 4, 5).

Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3) So total no of arrangement = 4 $\times$ 4 $\times$ 3 $\times$ 2 $\times$ 1 = 4.4!

$\therefore$ Total arrangement = 5!

+ 4.4!

= 216

Q14

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

A 346

B 140

C 196

D 280

Correct Answer

Option C

Solution

Case 1 : No of ways student can answer 10 questions =

{}^5{C_4} \times {}^8{C_6}

= 140 Case 2 : No of ways student can answer 10 questions =

{}^5{C_5} \times {}^8{C_5}

= 56 $\therefore$ Total ways = 140 + 56 = 196

Q15

These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:

A

N \le 100

B

100 < N \le 140

C

140 < N \le 190\,

D

N > 190

Correct Answer

Option A

Solution

We need 3 points to create a triangle. With 10 points number of triangle possible

{}^{10}{C_3}

Here 6 points are on the same line so we can't make any triangle with those 6 points. So subtract

{}^{6}{C_3}

. $\therefore$

N = {}^{10}{C_3} - {}^6{C_3}

= {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}

= 120 - 20

= 100

Q16

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

A

8.{}^6{C_4}.{}^7{C_4}

B

6.7.{}^8{C_4}

C

6.8.{}^7{C_4}

.

D

7.{}^6{C_4}.{}^8{C_4}

Correct Answer

Option D

Solution

This problem is solved using gap method.

As here no 'S' is adjacent to each other so we have to put them in the gap.

So first write all the letters other than 'S' such a way that there is a gap between two letters.

Given word is MISSISSIPPI.

Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time _M_I_I_I_I_P_P_ Those seven letters M, I, I, I, I, P, P can be arranged in

{{7!} \over {4!2!}}

ways Those seven letters creates 8 gaps and we have to choose 4 gaps from those 8 gaps to put those four 'S' letters.

This can be done

{}^8{C_4}

ways. After placing those four 'S' letters we can arrange them in

{{4!} \over {4!}}

ways. Therefore, required number of words

= {{7!} \over {4!2!}} \times {}^8{C_4} \times {{4!} \over {4!}}

= {{7\,.\,6!} \over {4!4!}} \times {}^8{C_4}

= 7\,.\,{}^6{C_4}\,.\,{}^8{C_4}

Q17

In a shop there are five types of ice-cream available. A child buys six ice-cream. Statement - 1: The number of different ways the child can buy the six ice-cream is

{}^{10}{C_5}

. Statement - 2: The number of different ways the child can buy the six ice-cream is equal to the number of different ways of arranging 6 A and 4 B's in a row.

A Statement - 1 is false, Statement - 2 is true

B Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1

C Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1

D Statement - 1 is true, Statement - 2 is false

Correct Answer

Option A

Solution

Note : n items can be distribute among p persons are

{}^{n + p - 1}{C_{p - 1}}

ways.

Here n = 6 ice-cream p = 5 types of ice-cream Each ice-cream belongs to one of the 5 ice-cream type.

So chosen 6 ice-crean can be divide into 5 types of ice-cream.

$\therefore$ The number of different ways the child can buy the six ice-cream is =

{}^{6 + 5 - 1}{C_{5 - 1}}

=

{}^{10}{C_4}

$\therefore$ Statement - 1 is false. Number of different ways of arranging 6 A and 4 B's in a row =

{{10!} \over {6!4!}} = {}^{10}{C_4}

$\therefore$ Statement - 2 is true.

Q18

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is :

A at least 500 but less than 750

B at least 750 but less than 1000

C at least 1000

D less than 500

Correct Answer

Option C

Solution

From 6 different novels 4 novels can be chosen =

{}^6{C_4}

ways And from 4 different dictionaries 1 can be chosen =

{}^3{C_1}

ways $\therefore$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen =

{}^6{C_4} \times {}^3{C_1}

ways Let 4 novels are N1, N2, N3, N4 and 1 dictionary is D1.

Dictionary should be in the middle.

So the arrangement will be like this _ _ D1 _ _ On those 4 blank places 4 novels N1, N2, N3, N4 can be placed.

And 4 novels can be arrange

4!

ways. $\therefore$ Total no of ways =

{}^6{C_4} \times {}^3{C_1}

\times 4!

= 1080

Q19

There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

A 36

B 66

C 108

D 3

Correct Answer

Option C

Solution

Thus number of ways

= ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108

Q20

Statement - 1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is

{}^9{C_3}

. Statement - 2: The number of ways of choosing any 3 places from 9 different places is

{}^9{C_3}

.

A Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1.

B Statement - 1 is true, Statement - 2 is false.

C Statement - 1 is false, Statement - 2 is true.

D Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1.

Correct Answer

Option A

Solution

Let XA, XB, XC and XD represent number of balls present in box A, B, C and D respectively.

As no box can be empty so, XA $\ge$ 1, XB $\ge$ 1, XC $\ge$ 1 and XD $\ge$ 1 $\Rightarrow$ XA $-$ 1 $\ge$ 0, $\Rightarrow$ XB $-$ 1 $\ge$ 0, $\Rightarrow$ XC $-$ 1 $\ge$ 0 and $\Rightarrow$ XD $-$ 1 $\ge$ 0 tA $\ge$ 0, tB $\ge$ 0, tC $\ge$ 0 and tD $\ge$ 0 According to the question, XA + XB + XC + XD = 10 $\Rightarrow$ (XA $-$ 1) + (XB $-$ 1) + (XC $-$ 1) + (XD $-$ 1) = 6 $\Rightarrow$ tA + tB + tC + tD = 6 Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6 From formula we know, n things can be distributed among r people in

{}^{n + r - 1}{C_{r - 1}}

ways where each people can have either 0 or more things. $\therefore$ 6 balls can be distributed among 4 boxes in

{}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}

ways where each box can have either 0 or more balls.

Therefore, Statement 1 is correct.

The number of ways of choosing any 3 places from 9 different places is

{}^9{C_3}

ways. But Statement - 2 is not the correct explanation of Statement - 1.