QBManager

Permutation and Combination

119306 Using the letters of the word TRICK a five letter word with distinct letters is formed such that $C$ is in the middle. In how many ways this is possible?

1 6

2 120

3 24

4 72

Explanation:

C Given the letters of the
Word
T R I C k
Total words = 5
original image
$= 4!$
$= 4 \times 3 \times 2 \times 1 = 24$
Total no of ways $= 24$

TS EAMCET-2017

Permutation and Combination

119308 Among the statements :
$(S_{1}) : 2023^{2022} - 1999^{2022}$ is divisible by 8
$(S_{2}) : 13 (13)^{n} - 11 n - 13$ is divisible by 8
Infinitely many $n \in N$ .

1 both

(S_{1})

and

(S_{2})

are incorrect

2 only

(S_{2})

is correct

3 both

(S_{1})

and

(S_{2})

are correct

4 only

(S_{1})

is correct

Explanation:

C Statements
(1) Given,
$S_{1} = 2023^{2022} - 1999^{2022}$
$S_{1} = (1999 + 24)^{2022} - (1999)^{2022}$
$^{2022} C_{1} (1999)^{2021} (24) +^{2022} C_{2} (1999)^{2020}$
$(24)^{2} + \dots \dots . . Soon$
$S_{1}$ is divisible by 8
Hence statement $I^{st}$ is correct
$S_{2} = 13 (13)^{n} - 11 n - 13$
$13^{n} = (1 + 12)^{n} = 1 + 12 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$
$13 (13^{n}) - 11 n - 13 = 145 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$ $∵$ If $(n = 144 m, m \in N)$ then
If is divisible by 144

JEE Main-06.04.2023

Permutation and Combination

119309 $\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to:

1

22! - 21

!

2

22! - 2 (21!)

3

21! - 2 (20

!)

4

21! - 20

!

Explanation:

B $\sum_{r = 1}^{20} (r^{2} + 1) r!$
$\sum_{r = 1}^{20} ((r + 1)^{2} - 2 r) r!$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r \cdot r!) - \sum_{r = 1}^{20} (r . r!)$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r . r!) - \sum_{r = 1}^{20} ((r + 1)! - r!)$
$= (21 . ∣ 21 - 1) - (∣ 21 - 1)$
$= 20.21! = 22! - 2.21!$

JEE Main-29.07.2022

Permutation and Combination

119310 The number of ways, in which 5 girls and 7 boys can be seated at round table so that no two girls sit together, is

1

126 (5!)^{2}

2

7 (360)^{2}

3 720

4

7 (720)^{2}

Explanation:

A Total number of ways
$= 6! \times^{7} C_{5} \times 5$ !
$= 6.5 .4 .3 .2 .1 \times \frac{7!}{5! (7 - 5)!} \times 5!$
$= 720 \times \frac{7.6 .5!}{5! 2!} \times 5!$
$= 2 \times 360 \times 7 \times 3 \times 120$
$= 126 \times (5!)^{2}$

JEE Main-08.04.2023

Permutation and Combination

119306 Using the letters of the word TRICK a five letter word with distinct letters is formed such that $C$ is in the middle. In how many ways this is possible?

1 6

2 120

3 24

4 72

Explanation:

C Given the letters of the
Word
T R I C k
Total words = 5
original image
$= 4!$
$= 4 \times 3 \times 2 \times 1 = 24$
Total no of ways $= 24$

TS EAMCET-2017

Permutation and Combination

119308 Among the statements :
$(S_{1}) : 2023^{2022} - 1999^{2022}$ is divisible by 8
$(S_{2}) : 13 (13)^{n} - 11 n - 13$ is divisible by 8
Infinitely many $n \in N$ .

1 both

(S_{1})

and

(S_{2})

are incorrect

2 only

(S_{2})

is correct

3 both

(S_{1})

and

(S_{2})

are correct

4 only

(S_{1})

is correct

Explanation:

C Statements
(1) Given,
$S_{1} = 2023^{2022} - 1999^{2022}$
$S_{1} = (1999 + 24)^{2022} - (1999)^{2022}$
$^{2022} C_{1} (1999)^{2021} (24) +^{2022} C_{2} (1999)^{2020}$
$(24)^{2} + \dots \dots . . Soon$
$S_{1}$ is divisible by 8
Hence statement $I^{st}$ is correct
$S_{2} = 13 (13)^{n} - 11 n - 13$
$13^{n} = (1 + 12)^{n} = 1 + 12 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$
$13 (13^{n}) - 11 n - 13 = 145 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$ $∵$ If $(n = 144 m, m \in N)$ then
If is divisible by 144

JEE Main-06.04.2023

Permutation and Combination

119309 $\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to:

1

22! - 21

!

2

22! - 2 (21!)

3

21! - 2 (20

!)

4

21! - 20

!

Explanation:

B $\sum_{r = 1}^{20} (r^{2} + 1) r!$
$\sum_{r = 1}^{20} ((r + 1)^{2} - 2 r) r!$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r \cdot r!) - \sum_{r = 1}^{20} (r . r!)$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r . r!) - \sum_{r = 1}^{20} ((r + 1)! - r!)$
$= (21 . ∣ 21 - 1) - (∣ 21 - 1)$
$= 20.21! = 22! - 2.21!$

JEE Main-29.07.2022

Permutation and Combination

119310 The number of ways, in which 5 girls and 7 boys can be seated at round table so that no two girls sit together, is

1

126 (5!)^{2}

2

7 (360)^{2}

3 720

4

7 (720)^{2}

Explanation:

A Total number of ways
$= 6! \times^{7} C_{5} \times 5$ !
$= 6.5 .4 .3 .2 .1 \times \frac{7!}{5! (7 - 5)!} \times 5!$
$= 720 \times \frac{7.6 .5!}{5! 2!} \times 5!$
$= 2 \times 360 \times 7 \times 3 \times 120$
$= 126 \times (5!)^{2}$

JEE Main-08.04.2023

Permutation and Combination

119306 Using the letters of the word TRICK a five letter word with distinct letters is formed such that $C$ is in the middle. In how many ways this is possible?

1 6

2 120

3 24

4 72

Explanation:

C Given the letters of the
Word
T R I C k
Total words = 5
original image
$= 4!$
$= 4 \times 3 \times 2 \times 1 = 24$
Total no of ways $= 24$

TS EAMCET-2017

Permutation and Combination

119308 Among the statements :
$(S_{1}) : 2023^{2022} - 1999^{2022}$ is divisible by 8
$(S_{2}) : 13 (13)^{n} - 11 n - 13$ is divisible by 8
Infinitely many $n \in N$ .

1 both

(S_{1})

and

(S_{2})

are incorrect

2 only

(S_{2})

is correct

3 both

(S_{1})

and

(S_{2})

are correct

4 only

(S_{1})

is correct

Explanation:

C Statements
(1) Given,
$S_{1} = 2023^{2022} - 1999^{2022}$
$S_{1} = (1999 + 24)^{2022} - (1999)^{2022}$
$^{2022} C_{1} (1999)^{2021} (24) +^{2022} C_{2} (1999)^{2020}$
$(24)^{2} + \dots \dots . . Soon$
$S_{1}$ is divisible by 8
Hence statement $I^{st}$ is correct
$S_{2} = 13 (13)^{n} - 11 n - 13$
$13^{n} = (1 + 12)^{n} = 1 + 12 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$
$13 (13^{n}) - 11 n - 13 = 145 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$ $∵$ If $(n = 144 m, m \in N)$ then
If is divisible by 144

JEE Main-06.04.2023

Permutation and Combination

119309 $\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to:

1

22! - 21

!

2

22! - 2 (21!)

3

21! - 2 (20

!)

4

21! - 20

!

Explanation:

B $\sum_{r = 1}^{20} (r^{2} + 1) r!$
$\sum_{r = 1}^{20} ((r + 1)^{2} - 2 r) r!$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r \cdot r!) - \sum_{r = 1}^{20} (r . r!)$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r . r!) - \sum_{r = 1}^{20} ((r + 1)! - r!)$
$= (21 . ∣ 21 - 1) - (∣ 21 - 1)$
$= 20.21! = 22! - 2.21!$

JEE Main-29.07.2022

Permutation and Combination

119310 The number of ways, in which 5 girls and 7 boys can be seated at round table so that no two girls sit together, is

1

126 (5!)^{2}

2

7 (360)^{2}

3 720

4

7 (720)^{2}

Explanation:

A Total number of ways
$= 6! \times^{7} C_{5} \times 5$ !
$= 6.5 .4 .3 .2 .1 \times \frac{7!}{5! (7 - 5)!} \times 5!$
$= 720 \times \frac{7.6 .5!}{5! 2!} \times 5!$
$= 2 \times 360 \times 7 \times 3 \times 120$
$= 126 \times (5!)^{2}$

JEE Main-08.04.2023

Permutation and Combination

119306 Using the letters of the word TRICK a five letter word with distinct letters is formed such that $C$ is in the middle. In how many ways this is possible?

1 6

2 120

3 24

4 72

Explanation:

C Given the letters of the
Word
T R I C k
Total words = 5
original image
$= 4!$
$= 4 \times 3 \times 2 \times 1 = 24$
Total no of ways $= 24$

TS EAMCET-2017

Permutation and Combination

119308 Among the statements :
$(S_{1}) : 2023^{2022} - 1999^{2022}$ is divisible by 8
$(S_{2}) : 13 (13)^{n} - 11 n - 13$ is divisible by 8
Infinitely many $n \in N$ .

1 both

(S_{1})

and

(S_{2})

are incorrect

2 only

(S_{2})

is correct

3 both

(S_{1})

and

(S_{2})

are correct

4 only

(S_{1})

is correct

Explanation:

C Statements
(1) Given,
$S_{1} = 2023^{2022} - 1999^{2022}$
$S_{1} = (1999 + 24)^{2022} - (1999)^{2022}$
$^{2022} C_{1} (1999)^{2021} (24) +^{2022} C_{2} (1999)^{2020}$
$(24)^{2} + \dots \dots . . Soon$
$S_{1}$ is divisible by 8
Hence statement $I^{st}$ is correct
$S_{2} = 13 (13)^{n} - 11 n - 13$
$13^{n} = (1 + 12)^{n} = 1 + 12 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$
$13 (13^{n}) - 11 n - 13 = 145 n +^{n} C_{2} \cdot 12^{2} +^{n} C_{3} \cdot 12^{3}$ $∵$ If $(n = 144 m, m \in N)$ then
If is divisible by 144

JEE Main-06.04.2023

Permutation and Combination

119309 $\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to:

1

22! - 21

!

2

22! - 2 (21!)

3

21! - 2 (20

!)

4

21! - 20

!

Explanation:

B $\sum_{r = 1}^{20} (r^{2} + 1) r!$
$\sum_{r = 1}^{20} ((r + 1)^{2} - 2 r) r!$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r \cdot r!) - \sum_{r = 1}^{20} (r . r!)$
$\sum_{r = 1}^{20} ((r + 1) (r + 1)! - r . r!) - \sum_{r = 1}^{20} ((r + 1)! - r!)$
$= (21 . ∣ 21 - 1) - (∣ 21 - 1)$
$= 20.21! = 22! - 2.21!$

JEE Main-29.07.2022

Permutation and Combination

119310 The number of ways, in which 5 girls and 7 boys can be seated at round table so that no two girls sit together, is

1

126 (5!)^{2}

2

7 (360)^{2}

3 720

4

7 (720)^{2}

Explanation:

A Total number of ways
$= 6! \times^{7} C_{5} \times 5$ !
$= 6.5 .4 .3 .2 .1 \times \frac{7!}{5! (7 - 5)!} \times 5!$
$= 720 \times \frac{7.6 .5!}{5! 2!} \times 5!$
$= 2 \times 360 \times 7 \times 3 \times 120$
$= 126 \times (5!)^{2}$

JEE Main-08.04.2023