QBManager

Permutation and Combination

119248 If $\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$ , then

1

b^{2} = a (b + c)

2

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1

3

a^{2} = c (a + b)

4

abc = 1

Explanation:

A We have,
$\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$
$\frac{1}{a} \frac{n!}{(n - r + 1)!} = \frac{1}{b} \frac{n!}{(n - r)!} = \frac{1}{c} \frac{n!}{(n - r - 1)!}$
$\Rightarrow \frac{b}{a} = n - r + 1$
$\frac{c}{b} = n - r$
$From equation (i) and (ii), we get-$
$\frac{b}{a} = \frac{c}{b} + 1 \Rightarrow b^{2} = a (b + c)$
From equation (i) and (ii), we get-

COMEDK-2019

Permutation and Combination

119249 If $\frac{n!}{5! (n - 1)!}$ and $\frac{n!}{7! (n - 3)!}$ are in the ratio 21 : 1 , then find the value of $n$ ,

1 1

2 2

3 3

4 4

Explanation:

C We have,
$\frac{n!}{5! (n - 1)!} : \frac{n!}{7! (n - 3)!} = 21 : 1$
$\Rightarrow \frac{n!}{5! (n - 1)!} \times \frac{7! (n - 3)!}{n!} = \frac{21}{1}$
$\Rightarrow \frac{7! (n - 3)!}{5! (n - 1)!} = \frac{21}{1}$
$\frac{7 \times 6 \times 5! (n - 3)!}{5! (n - 1) (n - 2) (n - 3)!} = \frac{21}{1}$
$\Rightarrow \frac{42}{n^{2} - 3 n + 2} = \frac{21}{1}$
$\Rightarrow n (n - 3) = 0 \Rightarrow n = 0 or n = 3$
$But when n = 0, then (n - 1)! and (n - 3)!are not defined.$
$Thus, n = 3$
But when $n = 0$ , then ( $n - 1$ )! and ( $n - 3)$ !are not defined. Thus, $n = 3$

COMEDK-2019

Permutation and Combination

119250 The number of integral solutions of $x + y + z =$ 0, with $x \geq - 5, y \geq - 5, z \geq - 5$ , is

1 135

2 136

3 455

4 105

Explanation:

B We have,
$x + y + z = 0$
Where, $x \geq - 5, y \geq - 5, z \geq - 5$
Now put :-
$x = t_{1} - 5, y = t_{2} - 5$ and $z = t_{3} - 5$
$∴ t_{1} + t_{2} + t_{3} = 15$ where $t_{1}, t_{2}, t_{3} \geq 0$
Required number of solutions -
$=^{15 + 3 - 1} C_{3 - 1}$
$=^{17} C_{2}$
$= \frac{17!}{2! 15!} = \frac{17 \times 16 \times 15!}{2 \times 1 \times 15!} = 136$

SRM JEEE-2008

Permutation and Combination

119251 The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is

1 24

2 52

3 48

4 16

Explanation:

D Required no. of triangles $=$ total no. of
triangles - no. of triangles having one side
common - no. of triangles having two sides
common.
$=^{8} C_{3} - 8 \times 4 - 8$
$= \frac{8!}{3! 5!} - 32 - 8 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} - 32 - 8$
$= 56 - 32 - 8 = 56 - 40 = 16$

SRM JEEE-2007

Permutation and Combination

119248 If $\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$ , then

1

b^{2} = a (b + c)

2

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1

3

a^{2} = c (a + b)

4

abc = 1

Explanation:

A We have,
$\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$
$\frac{1}{a} \frac{n!}{(n - r + 1)!} = \frac{1}{b} \frac{n!}{(n - r)!} = \frac{1}{c} \frac{n!}{(n - r - 1)!}$
$\Rightarrow \frac{b}{a} = n - r + 1$
$\frac{c}{b} = n - r$
$From equation (i) and (ii), we get-$
$\frac{b}{a} = \frac{c}{b} + 1 \Rightarrow b^{2} = a (b + c)$
From equation (i) and (ii), we get-

COMEDK-2019

Permutation and Combination

119249 If $\frac{n!}{5! (n - 1)!}$ and $\frac{n!}{7! (n - 3)!}$ are in the ratio 21 : 1 , then find the value of $n$ ,

1 1

2 2

3 3

4 4

Explanation:

C We have,
$\frac{n!}{5! (n - 1)!} : \frac{n!}{7! (n - 3)!} = 21 : 1$
$\Rightarrow \frac{n!}{5! (n - 1)!} \times \frac{7! (n - 3)!}{n!} = \frac{21}{1}$
$\Rightarrow \frac{7! (n - 3)!}{5! (n - 1)!} = \frac{21}{1}$
$\frac{7 \times 6 \times 5! (n - 3)!}{5! (n - 1) (n - 2) (n - 3)!} = \frac{21}{1}$
$\Rightarrow \frac{42}{n^{2} - 3 n + 2} = \frac{21}{1}$
$\Rightarrow n (n - 3) = 0 \Rightarrow n = 0 or n = 3$
$But when n = 0, then (n - 1)! and (n - 3)!are not defined.$
$Thus, n = 3$
But when $n = 0$ , then ( $n - 1$ )! and ( $n - 3)$ !are not defined. Thus, $n = 3$

COMEDK-2019

Permutation and Combination

119250 The number of integral solutions of $x + y + z =$ 0, with $x \geq - 5, y \geq - 5, z \geq - 5$ , is

1 135

2 136

3 455

4 105

Explanation:

B We have,
$x + y + z = 0$
Where, $x \geq - 5, y \geq - 5, z \geq - 5$
Now put :-
$x = t_{1} - 5, y = t_{2} - 5$ and $z = t_{3} - 5$
$∴ t_{1} + t_{2} + t_{3} = 15$ where $t_{1}, t_{2}, t_{3} \geq 0$
Required number of solutions -
$=^{15 + 3 - 1} C_{3 - 1}$
$=^{17} C_{2}$
$= \frac{17!}{2! 15!} = \frac{17 \times 16 \times 15!}{2 \times 1 \times 15!} = 136$

SRM JEEE-2008

Permutation and Combination

119251 The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is

1 24

2 52

3 48

4 16

Explanation:

D Required no. of triangles $=$ total no. of
triangles - no. of triangles having one side
common - no. of triangles having two sides
common.
$=^{8} C_{3} - 8 \times 4 - 8$
$= \frac{8!}{3! 5!} - 32 - 8 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} - 32 - 8$
$= 56 - 32 - 8 = 56 - 40 = 16$

SRM JEEE-2007

Permutation and Combination

119248 If $\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$ , then

1

b^{2} = a (b + c)

2

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1

3

a^{2} = c (a + b)

4

abc = 1

Explanation:

A We have,
$\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$
$\frac{1}{a} \frac{n!}{(n - r + 1)!} = \frac{1}{b} \frac{n!}{(n - r)!} = \frac{1}{c} \frac{n!}{(n - r - 1)!}$
$\Rightarrow \frac{b}{a} = n - r + 1$
$\frac{c}{b} = n - r$
$From equation (i) and (ii), we get-$
$\frac{b}{a} = \frac{c}{b} + 1 \Rightarrow b^{2} = a (b + c)$
From equation (i) and (ii), we get-

COMEDK-2019

Permutation and Combination

119249 If $\frac{n!}{5! (n - 1)!}$ and $\frac{n!}{7! (n - 3)!}$ are in the ratio 21 : 1 , then find the value of $n$ ,

1 1

2 2

3 3

4 4

Explanation:

C We have,
$\frac{n!}{5! (n - 1)!} : \frac{n!}{7! (n - 3)!} = 21 : 1$
$\Rightarrow \frac{n!}{5! (n - 1)!} \times \frac{7! (n - 3)!}{n!} = \frac{21}{1}$
$\Rightarrow \frac{7! (n - 3)!}{5! (n - 1)!} = \frac{21}{1}$
$\frac{7 \times 6 \times 5! (n - 3)!}{5! (n - 1) (n - 2) (n - 3)!} = \frac{21}{1}$
$\Rightarrow \frac{42}{n^{2} - 3 n + 2} = \frac{21}{1}$
$\Rightarrow n (n - 3) = 0 \Rightarrow n = 0 or n = 3$
$But when n = 0, then (n - 1)! and (n - 3)!are not defined.$
$Thus, n = 3$
But when $n = 0$ , then ( $n - 1$ )! and ( $n - 3)$ !are not defined. Thus, $n = 3$

COMEDK-2019

Permutation and Combination

119250 The number of integral solutions of $x + y + z =$ 0, with $x \geq - 5, y \geq - 5, z \geq - 5$ , is

1 135

2 136

3 455

4 105

Explanation:

B We have,
$x + y + z = 0$
Where, $x \geq - 5, y \geq - 5, z \geq - 5$
Now put :-
$x = t_{1} - 5, y = t_{2} - 5$ and $z = t_{3} - 5$
$∴ t_{1} + t_{2} + t_{3} = 15$ where $t_{1}, t_{2}, t_{3} \geq 0$
Required number of solutions -
$=^{15 + 3 - 1} C_{3 - 1}$
$=^{17} C_{2}$
$= \frac{17!}{2! 15!} = \frac{17 \times 16 \times 15!}{2 \times 1 \times 15!} = 136$

SRM JEEE-2008

Permutation and Combination

119251 The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is

1 24

2 52

3 48

4 16

Explanation:

D Required no. of triangles $=$ total no. of
triangles - no. of triangles having one side
common - no. of triangles having two sides
common.
$=^{8} C_{3} - 8 \times 4 - 8$
$= \frac{8!}{3! 5!} - 32 - 8 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} - 32 - 8$
$= 56 - 32 - 8 = 56 - 40 = 16$

SRM JEEE-2007

Permutation and Combination

119248 If $\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$ , then

1

b^{2} = a (b + c)

2

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1

3

a^{2} = c (a + b)

4

abc = 1

Explanation:

A We have,
$\frac{^{n} P_{r - 1}}{a} = \frac{^{n} P_{r}}{b} = \frac{^{n} P_{r + 1}}{c}$
$\frac{1}{a} \frac{n!}{(n - r + 1)!} = \frac{1}{b} \frac{n!}{(n - r)!} = \frac{1}{c} \frac{n!}{(n - r - 1)!}$
$\Rightarrow \frac{b}{a} = n - r + 1$
$\frac{c}{b} = n - r$
$From equation (i) and (ii), we get-$
$\frac{b}{a} = \frac{c}{b} + 1 \Rightarrow b^{2} = a (b + c)$
From equation (i) and (ii), we get-

COMEDK-2019

Permutation and Combination

119249 If $\frac{n!}{5! (n - 1)!}$ and $\frac{n!}{7! (n - 3)!}$ are in the ratio 21 : 1 , then find the value of $n$ ,

1 1

2 2

3 3

4 4

Explanation:

C We have,
$\frac{n!}{5! (n - 1)!} : \frac{n!}{7! (n - 3)!} = 21 : 1$
$\Rightarrow \frac{n!}{5! (n - 1)!} \times \frac{7! (n - 3)!}{n!} = \frac{21}{1}$
$\Rightarrow \frac{7! (n - 3)!}{5! (n - 1)!} = \frac{21}{1}$
$\frac{7 \times 6 \times 5! (n - 3)!}{5! (n - 1) (n - 2) (n - 3)!} = \frac{21}{1}$
$\Rightarrow \frac{42}{n^{2} - 3 n + 2} = \frac{21}{1}$
$\Rightarrow n (n - 3) = 0 \Rightarrow n = 0 or n = 3$
$But when n = 0, then (n - 1)! and (n - 3)!are not defined.$
$Thus, n = 3$
But when $n = 0$ , then ( $n - 1$ )! and ( $n - 3)$ !are not defined. Thus, $n = 3$

COMEDK-2019

Permutation and Combination

119250 The number of integral solutions of $x + y + z =$ 0, with $x \geq - 5, y \geq - 5, z \geq - 5$ , is

1 135

2 136

3 455

4 105

Explanation:

B We have,
$x + y + z = 0$
Where, $x \geq - 5, y \geq - 5, z \geq - 5$
Now put :-
$x = t_{1} - 5, y = t_{2} - 5$ and $z = t_{3} - 5$
$∴ t_{1} + t_{2} + t_{3} = 15$ where $t_{1}, t_{2}, t_{3} \geq 0$
Required number of solutions -
$=^{15 + 3 - 1} C_{3 - 1}$
$=^{17} C_{2}$
$= \frac{17!}{2! 15!} = \frac{17 \times 16 \times 15!}{2 \times 1 \times 15!} = 136$

SRM JEEE-2008

Permutation and Combination

119251 The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is

1 24

2 52

3 48

4 16

Explanation:

D Required no. of triangles $=$ total no. of
triangles - no. of triangles having one side
common - no. of triangles having two sides
common.
$=^{8} C_{3} - 8 \times 4 - 8$
$= \frac{8!}{3! 5!} - 32 - 8 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} - 32 - 8$
$= 56 - 32 - 8 = 56 - 40 = 16$

SRM JEEE-2007