Permutations and Combinations — Page 8 | JEE Mathematics

Q71

If

\mathrm{n}

is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then

\mathrm{n}

is equal to :

A 47

B 53

C 51

D 43

Correct Answer

Option C

Solution

To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.

The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 indistinguishable offices) where each part represents the number of employees in each office.

The partitions of 5 into at most 4 parts are as follows : $(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.

$(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.

$(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.

$(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.

$(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.

$(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.

However, since the offices are indistinguishable, we must not count partitions that differ only by the order of the parts.

This means that partitions like $(5, 0, 0, 0)$ , $(0, 5, 0, 0)$ , $(0, 0, 5, 0)$ , and $(0, 0, 0, 5)$ all represent the same scenario and thus are counted as one.

So we end up with 6 distinct partition scenarios.

Now, for each partition, we need to find the number of ways to assign the employees according to each partition : $(5, 0, 0, 0)$ – All employees are in one office.

There is 1 way to do this because the offices are indistinguishable.

$(4, 1, 0, 0)$ - For the group with 4 employees, there are

{}^5{C_4}

ways to choose which 4 out of the 5 will be together.

Then the remaining employee is automatically assigned to the second office.

So, there are

{}^5{C_4}

ways for this partition.

$(3, 2, 0, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways to choose them, and for the group with 2 employees, there are

{}^2{C_2}

ways to choose them (which is essentially 1 way since the last 2 employees are automatically grouped).

But the 2 offices these groups can occupy are indistinguishable, so we do not multiply by 2.

So, there are

{}^5{C_3}

ways for this partition. $(3, 1, 1, 0)$ - For the group with 3 employees, there are

{}^5{C_3}

ways.

Then we have two indistinguishable offices, each with 1 employee.

The order does not matter as offices are indistinguishable.

So, there are simply

{}^5{C_3}

ways. $(2, 2, 1, 0)$ – For the first group of 2 employees, there are

{}^5{C_2}

ways to choose them. For the next group of 2 employees, there are

{}^3{C_2}

ways from the remaining 3.

Then, the last employee is alone, and since the offices are indistinguishable, there are no further combinations to consider.

However, we have double-counted since the two groups of two are indistinguishable.

To adjust for this, we divide by 2.

This gives us $\dfrac{{}^5{C_2} \cdot {}^3{C_2}}{2}$ ways.

$(2, 1, 1, 1)$ - For the group with 2 employees, there are ${}^5{C_2}$ ways to choose them.

The other three employees are each in their own office, with no further combinations since the offices are indistinguishable, so there are ${}^5{C_2}$ ways for this partition.

Let's compute these cases : $(5, 0, 0, 0)$ corresponds to $1$ way. $(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways. $(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways. $(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways. $(2, 2, 1, 0)$ corresponds to $\dfrac{{}^5{C_2} \cdot {}^3{C_2}}{2} = \dfrac{10 \cdot 3}{2} = 15$ ways. $(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.

Adding up all the ways we get :

1 + 5 + 10 + 10 + 15 + 10 = 51

Therefore, the number of ways five different employees can sit into four indistinguishable offices ( $\mathrm{n}$ ) is 51, which corresponds to Option C.

Q72

Let

\alpha=\dfrac{(4 !) !}{(4 !)^{3 !}}

and

\beta=\dfrac{(5 !) !}{(5 !)^{4 !}}

. Then :

A

\alpha \in \mathbf{N}

and

\beta \in \mathbf{N}

B

\alpha \in \mathbf{N}

and

\beta \notin \mathbf{N}

C

\alpha \notin \mathbf{N}

and

\beta \in \mathbf{N}

D

\alpha \notin \mathbf{N}

and

\beta \notin \mathbf{N}

Correct Answer

Option A

Solution

\begin{aligned} & \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\ & \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}} \end{aligned}

Let 24 distinct objects are divided into 6 groups of 4 objects in each group. No. of ways of formation of group

=\frac{24 !}{(4 !)^6 .6 !} \in \mathrm{N}

Similarly, Let 120 distinct objects are divided into 24 groups of 5 objects in each group.

No. of ways of formation of groups

=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in \mathrm{N}

Q73

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

A 130

B 136

C 142

D 406

Correct Answer

Option B

Solution

To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls and bins"), which is a way to solve problems involving distributing identical items into distinct groups with certain restrictions.

First, since each child must get at least 2 apples, let's give 2 apples to each child right away.

That accounts for 6 apples (2 apples for each of the 3 children).

Now, we have 21 - 6 = 15 apples left to distribute freely among the three children.

The "stars and bars" technique involves representing the apples as stars (*) and the divisions between children as bars (|).

For example, if we had 5 apples to distribute among three children, one possible distribution could be represented as **|*|**.

This means the first child gets 2 apples, the second child gets 1 apple, and the third child gets 2 apples.

In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the partitions.

We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars corresponds to a distribution of the apples.

The total number of objects we're arranging is 15 apples + 2 bars = 17 objects.

We need to choose 2 positions out of these 17 to place the bars.

The remaining positions will be occupied by the stars (apples).

The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient:

\text{Number of ways} = \binom{17}{2} = \frac{17!}{2!(17-2)!} = \frac{17 \times 16}{2 \times 1} = 136

Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child gets at least 2 apples.

The correct answer is Option B: 136.

Q74

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to:

A 179

B 177

C 175

D 181

Correct Answer

Option A

Solution

\begin{aligned} & 2 M \\ & 2 A \\ & 2 T \\ & H, E, I, C, S \end{aligned}

Case-I 2 Alike 2 Alike 1 Diff

{ }^3 C_2 \times{ }^6 C_1=18

Case-II 2 Alike + 3 Diff

{ }^3 C_1 \times{ }^7 C_3=105

Case-III All different

{ }^8 C_5=56

Total ways

=179

Q75

Let

[t]

be the greatest integer less than or equal to

t

. Let

A

be the set of all prime factors of 2310 and

f: A \rightarrow \mathbb{Z}

be the function

f(x)=\left[\log _2\left(x^2+\left[\dfrac{x^3}{5}\right]\right)\right]

. The number of one-to-one functions from

A

to the range of

f

is

A 20

B 120

C 25

D 24

Correct Answer

Option B

Solution

\begin{aligned} & A=\{2,3,5,7,11\} \\ & f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right] \\ & \text{ Ranges } f(x)=\{2,3,5,6,8\} \end{aligned}

\text{ Number of one-one } A \rightarrow R_f

5 \times 4 \times 3 \times 2 \times 1=120

Q76

Let the set

S=\{2,4,8,16, \ldots, 512\}

be partitioned into 3 sets

A, B, C

with equal number of elements such that

\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}

and

\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{A} \cap \mathrm{C}=\phi

. The maximum number of such possible partitions of

S

is equal to:

A 1640

B 1520

C 1710

D 1680

Correct Answer

Option D

Solution

Given set

S=\left\{2^1, 2^2, \ldots 2^9\right\}

which consist of 9 elements. Maximum number of possible partitions (in set

A, B

and

C

)

={ }^9 C_3 \cdot{ }^6 C_3 \cdot{ }^3 C_3=1680

Q77

60 words can be made using all the letters of the word

\mathrm{BHBJO}

, with or without meaning. If these words are written as in a dictionary, then the

50^{\text{th }}

word is:

A OBBJH

B HBBJO

C OBBHJ

D JBBOH

Correct Answer

Option A

Solution

To find the

50^{\text{th}}

word formed by the letters of "BHBJO" as if listed in a dictionary, let's analyze the arrangement methodically.

Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.

First, calculate the total number of permutations of these letters:

\frac{5!}{2!} = 60

Let's arrange the letters in alphabetical order first: B, B, H, J, O.

We need to systematically count the words while following dictionary order: 1.

Words starting with B: Next position letters: B, H, J, O Number of permutations:

\frac{4!}{1!} = 24

words Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically.

2.

Words starting with H: Next position letters: B, B, J, O Number of permutations:

\frac{4!}{2!} = 12

words After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more.

3.

Words starting with J: Next position letters: B, B, H, O Number of permutations:

\frac{4!}{2!} = 12

words Now tally is 36 + 12 = 48 words.

Still 2 more to reach 50.

4.

Words starting with O: Next position letters: B, B, H, J Number of permutations:

\frac{4!}{2!} = 12

words Our answer must be here since 48 + 2 more = 50 total. First permutation:

OBB H J

Second permutation (50th word):

OBB J H

So, the

50^{\text{th}}

word is: OBBJH Thus, the correct answer is Option A: OBBJH.

Q78

If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at

315^{\text{th }}

position in this arrangement is :

A NRAPUG

B NRAGUP

C NRAPGU

D NRAGPU

Correct Answer

Option C

Solution

NAGPUR Word at

315^{\text{th }}

position

\begin{aligned} & \text{ A...... }=5!=120 \\ & \text{ G....... }=5!=120 \\ & \text{ NA..... }=4!=24 \\ & \text{ NG..... }=4!=24 \\ & \text{ NP..... }=4!=24 \end{aligned}

..... Till 312 words

313^{\text{th }} \text{ word }=\text{ NRAGPU }

314^{\text{th }}

word $=$

\mathrm{NRAGUP}

315^{\text{th }} \text{ word }=\text{ NRAPGU }

Q79

Let

0 \leq r \leq n

. If

{ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=55: 35: 21

, then

2 n+5 r

is equal to :

A 62

B 60

C 55

D 50

Correct Answer

Option D

Solution

Given $0 \leq r \leq n$ .

If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$ , then we are to determine the value of $2n + 5r$ .

Step-by-Step Solution: Write the given proportions involving binomial coefficients: $\dfrac{n+1}{r+1} \times \binom{n}{r} : \binom{n}{r} : \dfrac{r}{n} \times \binom{n}{r} = 55 : 35 : 21$ Simplify the proportions: $\dfrac{n+1}{r+1} = \dfrac{55}{35} \quad \text{and} \quad \dfrac{n}{r} = \dfrac{35}{21}$ From the simplified ratios, we establish the following two equations: $\dfrac{n+1}{r+1} = \dfrac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad \text{.... (1)}$ $\dfrac{n}{r} = \dfrac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad \text{.... (2)}$ Solve equations (1) and (2) simultaneously: From equation (2), solve for $n$ : $3n = 5r \quad \Rightarrow \quad n = \dfrac{5r}{3}$ Substitute $n$ into equation (1): $7 \left(\dfrac{5r}{3}\right) - 11r = 4 \quad \Rightarrow \quad \dfrac{35r}{3} - 11r = 4$ $\dfrac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \dfrac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6$ Substituting $r$ back to find $n$ : $n = \dfrac{5 \times 6}{3} = 10$ Compute $2n + 5r$ : $2n + 5r = 2 \times 10 + 5 \times 6 = 20 + 30 = 50$ Thus, the value of $2n + 5r$ is 50.

Q80

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is

A 56

B 16

C 24

D 48

Correct Answer

Option B

Solution

To solve this problem, we need to determine the number of triangles formed by the vertices of a regular octagon such that none of the sides of the triangle is also a side of the octagon.

Let's start by counting the total number of triangles that can be formed using the 8 vertices of the octagon.

The number of ways to choose 3 vertices out of 8 is given by the combination formula:

\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

Next, we need to exclude those triangles that have at least one side coinciding with a side of the octagon.

Let's analyze how many such invalid triangles there can be.

Consider each side of the octagon.

For any given side, there are exactly 5 other vertices remaining (since we must exclude the two vertices that form the current side).

Out of these 5 vertices, we can choose any 1 to form a triangle that has one side common with the octagon.

Hence, for each side of the octagon, there are 5 such triangles.

Since the octagon has 8 sides, the total number of triangles that have at least one side as a side of the octagon is:

8 \times 5 = 40

Therefore, the number of triangles whose sides do not coincide with any sides of the octagon is:

56 - 40 = 16

So, the correct answer is: Option B: 16