Permutations and Combinations — Page 7 | JEE Mathematics

Q61

All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is :

A 324

B 328

C 326

D 327

Correct Answer

Option D

Solution

The letters of "MONDAY" arranged in alphabetical order are: A, D, M, N, O, Y.

Fix A as the first letter, we can arrange the remaining 5 letters in 5!

ways = 120 ways.

Fix D as the first letter, we can arrange the remaining 5 letters in 5!

ways = 120 ways.

So far, we have 120 (for A) + 120 (for D) = 240 words.

Now, let's proceed with words starting with M: Fix MA as the first two letters, we can arrange the remaining 4 letters in 4!

ways = 24 ways.

Fix MD as the first two letters, we can arrange the remaining 4 letters in 4!

ways = 24 ways.

Fix MN as the first two letters, we can arrange the remaining 4 letters in 4!

ways = 24 ways.

Now, we have 240 (for A and D) + 24 (for MA) + 24 (for MD) + 24 (for MN) = 312 words.

After MN, we consider words that start with MO: Fix MOA as the first three letters, we can arrange the remaining 3 letters in 3!

ways = 6 ways.

Fix MOD as the first three letters, we can arrange the remaining 3 letters in 3!

ways = 6 ways.

Adding these to the total, we have 312 (previous total) + 6 (for MOA) + 6 (for MOD) = 324 words.

Next, we consider words that start with 'MON'.

The word 'MONDAY' comes after 'MONAD' in dictionary order, so: Fix 'MONA' as the first four letters.

The remaining 2 letters ('D' and 'Y') can be arranged in 2!

ways, which gives us 2 more words: 'MONADY' and 'MONAYD'.

So, we have 324 (previous total) + 2 (for 'MONADY' and 'MONAYD') = 326 words.

Finally, we have the word 'MONDAY' itself, which is the 327th word.

Q62

If for some

m, n ;{ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3

and

{ }^{n-1} P_3:{ }^n P_4=1: 8

, then

{ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m

is equal to

A 380

B 376

C 372

D 384

Correct Answer

Option C

Solution

\begin{aligned} & { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\ & { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\ & { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\ & \therefore \mathrm{m}=2 \\ & \text{ And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 8 \\ & \frac{(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=\frac{1}{8} \\ & \therefore \mathrm{n}=8 \\ & \therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2 \\ & =8 \times 7 \times 6+\frac{9 \times 8}{2} \\ & =372 \end{aligned}

Q63

The number of triplets

(x, \mathrm{y}, \mathrm{z})

, where

x, \mathrm{y}, \mathrm{z}

are distinct non negative integers satisfying

x+y+z=15

, is :

A 136

B 80

C 92

D 114

Correct Answer

Option D

Solution

We have, $x+y+z=15$

\begin{aligned} \text{ Total number of solution } & ={ }^{15+3-1} C_{3-1} \\\\ & ={ }^{17} C_2=\frac{17 \times 16}{1 \times 2}=136 \end{aligned}

Now, we need to exclude the solutions where two of $(x, y, z)$ are the same.

1) For the case $x = y \neq z$ : $2x + z = 15$ The solutions are : $x = 0, z = 15$ $x = 1, z = 13$ $x = 2, z = 11$ $x = 3, z = 9$ $x = 4, z = 7$ $x = 5, z = 5$ (Not valid as all are the same) $x = 6, z = 3$ $x = 7, z = 1$ Out of these, 7 are valid.

Similarly, for the cases $y = z \neq x$ and $z = x \neq y$ , there will be 7 valid solutions for each, so a total of $7 \times 3 = 21$ solutions where two of the variables are equal.

Thus, the number of triplets where all are distinct is : $136 - 21 = 115$ There is one solution in which $\mathrm{x}=\mathrm{y}=\mathrm{z}$ Required answer $=136-21-1=114$

Q64

Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways, in which they can be transported, is :

A 560

B 1680

C 3360

D 1120

Correct Answer

Option B

Solution

Let $C_1, C_2$ and $C_3$ be the three cars in which 8 person are to be transported from city $A$ to city $B$ .

Let the number of persons transported in cars $C_1, C_2$ and $C_3$ are 3,3 and 2 respectively.

There are total $\dfrac{8 !}{3 ! 3 ! 2 ! 2 !}$ group $\therefore$ They can travel $\dfrac{8 !}{3 ! 3 ! 2 ! 2 !} \times 3$ !

ways or 1680 ways

Q65

If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which

\mathrm{C}

and

\mathrm{S}

do not come together, is

(6 !) \mathrm{k}

, then

\mathrm{k}

is equal to :

A 5670

B 1890

C 2835

D 945

Correct Answer

Option A

Solution

\text{ Total number of words }=\frac{11 !}{2 ! 2 ! 2 !}

Number of words in which $\mathrm{C}$ and $\mathrm{S}$ are together

=\frac{10 !}{2 ! 2 ! 2 !} \times 2 \text{ ! }

So, required number of words

\begin{aligned} & =\frac{11 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\ & =\frac{11 \times 10 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\ & =\frac{10 !}{2 ! 2 !}\left[\frac{11}{2}-1\right]=\frac{10 !}{2 ! 2 !} \times \frac{9}{2} \\\\ & =5670 \times 6 ! \\\\ & \Rightarrow k(6 !)=5670 \times 6 ! \\\\ & \Rightarrow k=5670 \end{aligned}

Q66

The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is :

A 16800

B 14800

C 18000

D 33600

Correct Answer

Option A

Solution

In the given word, vowels are : I, E, E, E, E Consonants are : N, D, P, N, D, N, C So, number of words $=\dfrac{8 !}{3 ! 2 !} \times \dfrac{5 !}{4 !}$ $=\dfrac{8 \times 7 \times 6 \times 5 \times 4}{2} \times 5=16800$ Concept : Out of $n$ objects, if $r$ things are same, so number of ways $=\dfrac{n !}{r !}$

Q67

The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together, is :

A 720

B

7(360)^{2}

C

7(720)^{2}

D

126(5 !)^{2}

Correct Answer

Option D

Solution

We have, Number of girls $=5$ Number of boys $=7$ So, number of ways of arranging boys around the table $=6$ !

and 5 girls can be arranged in 7 gaps in ${ }^7 \mathrm{P}_5$ ways.

$\therefore$ Required no. of ways $=6 ! \times{ }^7 \mathrm{P}_5$ $=126 \times(5 !)^2$

Q68

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

A 18

B 16

C 12

D 15

Correct Answer

Option D

Solution

3 \text{ Shelf empty: }(8,0,0,0) \rightarrow 1 \text{ way }

\left.2 \text{ shelf empty: } \begin{array}{c} (7,1,0,0) \\ (6,2,0,0) \\ (5,3,0,0) \\ (4,4,0,0) \end{array}\right] \rightarrow 4 \text{ ways }

\left.1 \text{ shelf empty: } \begin{array}{cc} (6,1,1,0) & (3,3,2,0) \\ (4,2,1,0) & (4,2,2,0) \\ (4,3,0) & \end{array}\right] \rightarrow 5 \text{ ways }

\left.0 \text{ Shelf empty : } \begin{array}{ll} (1,2,3,2) & (5,1,1,1) \\ (2,2,2,2) \\ (3,3,2,1,1) \\ (4,2,1,1) & \end{array}\right] \rightarrow 5 \text{ ways }

Total

=15

ways

Q69

Let the number of elements in sets

A

and

B

be five and two respectively. Then the number of subsets of

A \times B

each having at least 3 and at most 6 elements is :

A 782

B 772

C 752

D 792

Correct Answer

Option D

Solution

First, let's determine the number of elements in the Cartesian product

A \times B

. If set

A

has 5 elements and set

B

has 2 elements, then the number of elements in

A \times B

is:

|A \times B| = |A| \times |B| = 5 \times 2 = 10

We need to find the number of subsets of

A \times B

with at least 3 elements and at most 6 elements. The total number of elements in

A \times B

is 10, so we must consider the subsets containing 3, 4, 5, and 6 elements. The number of ways to choose

r

elements from a set of 10 elements is given by the binomial coefficient:

\binom{10}{r} = \frac{10!}{r!(10-r)!}

We will find the sum of the binomial coefficients for

r = 3, 4, 5, 6

: First,

\binom{10}{3}

:

\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

Next,

\binom{10}{4}

:

\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

Next,

\binom{10}{5}

:

\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252

Next,

\binom{10}{6}

:

\binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

Finally, we sum these values:

\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} = 120 + 210 + 252 + 210 = 792

Therefore, the number of subsets of

A \times B

each having at least 3 and at most 6 elements is: Option D - 792

Q70

All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is :

A 578

B 576

C 580

D 582

Correct Answer

Option D

Solution

Given word is PUBLIC.

Alphabetical order of letters is BCILPU.

So, number of words start with letter.

B - - - - is $5 !=120$ C ---- is $5 !=120$ I ----- is $5 !=120$ $\mathrm{L}-----$ is $5 !=120$ $\mathrm{PB}----$ is $4 !=24$ $\mathrm{PC}----$ is $4 !=24$ PI ---- is $4 !=24$ $\mathrm{PL}----$ is $4 !=24$ PUBC -- is $2 !=2$ PUBI - is $2 !=2$ PUBLC - is $1 !=1$ PUBLIC is $0 !=1$ $\therefore$ Serial number of the word PUBLIC $=4 \times 120+4 \times 24+2 \times 2+2 \times 1=582$