Permutations and Combinations — Page 9 | JEE Mathematics

Q81

Let

P

be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in

P

are formed by using the digits 1, 2 and 3 only, then the number of elements in the set

P

is :

A 164

B 158

C 161

D 173

Correct Answer

Option C

Solution

(i) number of numbers created using

1111133=\frac{7!}{5!2!} \Rightarrow 21

(ii) number of numbers created using

1111223=\frac{7!}{4!2!} \Rightarrow 105

(iii) number of numbers created using

\begin{aligned} & 1112222=\frac{7!}{4!3!} \Rightarrow 35 \\ & \text{ Total }=161 \end{aligned}

Q82

The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :

A 360

B 2520

C 1820

D 45

Correct Answer

Option B

Solution

To find the number of sequences of ten terms, each being either 0, 1, or 2, containing exactly five 1s and exactly three 2s, follow these steps: Determine the Remaining Terms: Since there are 5 ones and 3 twos, you will need 2 zeros to fill the sequence (because $5 + 3 + 2 = 10$ ).

Calculate the Number of Arrangements: You have a total of 10 positions to fill with these numbers (5 ones, 3 twos, 2 zeros).

The formula for calculating permutations of a multiset is: $\dfrac{10!}{5! \times 3! \times 2!}$ Where: $10!$ is the factorial of the total number of terms. $5!$ is the factorial for the number of 1s. $3!$ is the factorial for the number of 2s. $2!$ is the factorial for the number of 0s.

Result: Simplifying the calculation gives: $\dfrac{10!}{5! \times 3! \times 2!} = 2520$ Thus, there are 2520 possible sequences with the given conditions.

Q83

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is :

A 6084

B 5148

C 14950

D 4356

Correct Answer

Option B

Solution

First, note that we are choosing 5 distinct letters (in strictly increasing alphabetical order) such that the middle (third) letter is ‘M’.

Symbolically, if we denote the chosen letters as: $L_1 we want$ L_3 = \text{M} $. The English alphabet has 26 letters, and M is the$ 13^\text{th} $. Step 1: Letters before M The letters before M are$ \{A, B, C, \ldots, L\} $. There are 12 letters here ($ A $through$ L $). We need to pick 2 of these 12 letters to occupy$ L_1 $and$ L_2 $. The number of ways to choose 2 letters out of 12 is$ { }^{12} \mathrm{C}_2 $. Step 2: Letters after M The letters after M are$ \{N, O, P, \ldots, Z\} $. There are 13 letters here ($ N $through$ Z $). We need to pick 2 of these 13 letters to occupy$ L_4 $and$ L_5 $. The number of ways to choose 2 letters out of 13 is$ { }^{13} \mathrm{C}_2 $. Step 3: Multiply the choices Since these choices are independent (picking the two letters before M and two letters after M), the total number of ways is:$ { }^{12} \mathrm{C}_2 \;\times\;{ }^{13} \mathrm{C}_2 $Calculate each combination:$ { }^{12} \mathrm{C}_2 = \frac{12 \times 11}{2} = 66, \quad { }^{13} \mathrm{C}_2 = \frac{13 \times 12}{2} = 78. $So,$ { }^{12} \mathrm{C}_2 \times { }^{13} \mathrm{C}_2 = 66 \times 78 = 5148.

$ Answer: 5148 (Option B)

Q84

In a group of 3 girls and 4 boys, there are two boys

B_1

and

B_2

. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but

B_1

and

B_2

are not adjacent to each other, is :

A 120

B 96

C 72

D 144

Correct Answer

Option D

Solution

Let's break the problem down step by step: There are two blocks because all girls must stand together and all boys must stand together.

The two blocks can be arranged in:

2 \text{ ways} \quad \text{(i.e., girls first then boys, or boys first then girls).}

The girls can be arranged among themselves in:

3! = 6 \text{ ways.}

For the boys (4 in total), they must be arranged such that the specific boys $B_1$ and $B_2$ are not adjacent.

First, calculate the total number of arrangements of 4 boys:

4! = 24.

Next, count the arrangements where $B_1$ and $B_2$ are adjacent.

Think of $B_1$ and $B_2$ as a single unit.

This unit can be arranged in:

2! = 2 \text{ ways (since }B_1\text{ and }B_2\text{ can swap positions).}

Now, with this new unit, we have 3 units in total (the $B_1B_2$ unit and the other 2 boys), which can be arranged in:

3! = 6 \text{ ways.}

So, the number of arrangements where $B_1$ and $B_2$ are adjacent is:

2! \times 3! = 2 \times 6 = 12.

Therefore, the number of valid arrangements for the boys where $B_1$ and $B_2$ are not adjacent is:

24 - 12 = 12.

Finally, multiply all the factors together:

\text{Total ways} = 2 \times 6 \times 12 = 144.

Thus, the number of ways in which the girls and boys can stand in the queue under the given conditions is

\boxed{144}.

This corresponds to Option D.

Q85

The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is :

A 34000

B 37000

C 35000

D 36000

Correct Answer

Option D

Solution

DAUGHTER Total words $=8$ !

Total words in which vowels are together $=6!\times 3!$ words in which all vowels are not together

\begin{aligned} & =8!-6!\times 3! \\ & =6![56-6] \\ & =720 \times 50 \\ & =36000 \end{aligned}

Q86

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group

A

and the remaining 3 from group

B

, is equal to :

A 8925

B 9100

C 8575

D 8750

Correct Answer

Option A

Solution

\begin{array}{ll} \text{ C-I } & (3 \mathrm{G} \& 2 \mathrm{~B}) \&(1 \mathrm{G} \& 2 \mathrm{~B}) \\ \text{ C-II } & (2 \mathrm{G} \& 3 \mathrm{~B}) \&(2 \mathrm{G} \& 1 \mathrm{~B}) \\ \text{ C-III } & (1 \mathrm{G} \& 4 \mathrm{~B}) \&(3 \mathrm{G} \& 0 \mathrm{~B}) \end{array}

\begin{aligned} & \text{ Total }=\text{ C-I }+ \text{ C-II }+ \text{ C-III } \\ & ={ }^7 \mathrm{C}_2 \cdot{ }^3 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_1+{ }^7 \mathrm{C}_3 \cdot{ }^3 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_1{ }^5 \mathrm{C}_2+{ }^7 \mathrm{C}_4 \cdot{ }^3 \mathrm{C}_1 \cdot{ }^6 \mathrm{C}_0 \cdot{ }^5 \mathrm{C}_3 \\ & =8925 \end{aligned}

Q87

Let

{ }^n C_{r-1}=28,{ }^n C_r=56

and

{ }^n C_{r+1}=70

. Let

A(4 \operatorname{cost}, 4 \sin t), B(2 \sin t,-2 \cos t)

and

C\left(3 r-n, r^2-n-1\right)

be the vertices of a triangle

A B C

, where

t

is a parameter. If

(3 x-1)^2+(3 y)^2

=\alpha

, is the locus of the centroid of triangle ABC , then

\alpha

equals

A 18

B 8

C 20

D 6

Correct Answer

Option C

Solution

\begin{aligned} & { }^n C_{r-1}=28,{ }^n C_r=56 \\ & \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{28}{56} \\ & \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2} \\ & \frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2} \\ & 3 \mathrm{r}=\mathrm{n}+1\quad\text{..... (i)} \end{aligned}

\begin{aligned} & \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \\ & \frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5\quad\text{.... (ii)} \end{aligned}

\begin{aligned} & \text{ By (i) & (ii) } \\ & \begin{array}{l} (r=3),(n=8) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \operatorname{cost}) C(1,0) \\ (3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \sin t)^2+(4 \sin t-\operatorname{cost})^2 \\ (3 x-1)^2+(3 y)^2=20 \quad \therefore \text{ option }(1) \end{array} \end{aligned}

Q88

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0 ,

1,2,3,4,5,6,7

, such that the sum of their first and last digits should not be more than 8 , is

A 5720

B 5719

C 4608

D 4607

Correct Answer

Option D

Solution

Case I $5{ }_{---} 0$ Case II $5{ }_{---} 1$

\begin{array}{ll} 5 & 2 \\ 5 & 3 \\ 6 & 0 \\ 6 & 1 \\ 6 & 2 \\ 7 & 0 \end{array}

Case IX $7{ }_{---} 1$ $9 \times(8 \times 8 \times 8)=4608$ but 50000 is not included, so total numbers $4608-1=4607$

Q89

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

A 230

B 210

C 200

D 220

Correct Answer

Option B

Solution

To count the number of distinct triangles: Start with all possible triples of points.

From 12 points, the number of ways to choose any 3 is $\binom{12}{3}= \dfrac{12\cdot11\cdot10}{3\cdot2\cdot1}=220.$ Subtract the “invalid” triples that are collinear.

The only collinear sets of three points arise from the single line containing the 5 collinear points.

Number of ways to pick 3 points from those 5 is $\binom{5}{3}=10.$ Valid triangles = total triples − collinear triples.

$220-10 = 210.$ Hence, the total number of triangles that can be formed is 210.

Correct option: B

Q90

Line

L_1

of slope 2 and line

L_2

of slope

\dfrac{1}{2}

intersect at the origin O . In the first quadrant,

\mathrm{P}_1

,

P_2, \ldots, P_{12}

are 12 points on line

L_1

and

Q_1, Q_2, \ldots, Q_9

are 9 points on line

L_2

. Then the total number of triangles, that can be formed having vertices at three of the 22 points

\mathrm{O}, \mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{12}

,

\mathrm{Q}_1, \mathrm{Q}_2, \ldots, \mathrm{Q}_9

, is:

A 1026

B 1188

C 1134

D 1080

Correct Answer

Option C

Solution

Total triangles (2 points as $y=\dfrac{x}{2}, 1$ point on $y=\dfrac{x}{2}$ ) $+2\left(\right.$ points and $y=\dfrac{x}{2}, 1$ point on $y=2 x$ ) +(1 point on $y=2 x, 1)$ point on $y=\dfrac{x}{2}$ and origin)

={ }^9 C_2,{ }^{12} C_1+{ }^9 C_1{ }^{12} C_2+{ }^9 C_1 \cdot{ }^{12} C_1 \cdot{ }^1 C_1

=1134