QBManager

Permutation and Combination

119008 3 persons have 4 coats, 5 waist coats and 6 hats. The number of ways in which they put on the clothes are :

1

4^{3} \times 5^{3} \times 6^{3}

2

4 \times 5 \times 6

3

4! 5! 6

!

4 none of these

Explanation:

A : Given that,
3 person have 4 coats, 5 waist coat and 6 hats
Number of ways in which 3 men can wear 4 coats $=^{4} P_{3}$
Number of ways in which 3 men can bear 5 waistcoats
$=^{5} P_{3}$
Number of ways in which 3 men can bear 6 hats $=^{6} P_{3}$
$∴$ The total number of ways $=^{4} P_{3} \times^{5} P_{3} \times^{6} P_{3}$
$= \frac{4!}{1!} \times \frac{5!}{2!} \times \frac{6!}{3!}$
$= 24 \times 60 \times 120 = 172800$ .

BCECE-2005

Permutation and Combination

119009 A man has 7 relatives, 4 women and 3 men, His wife also has 7 relatives, 3 women and 4 men. What is the number of ways can they invite 3 women and 3 men so that 3 of them are the man's relatives and 3 of them are his wife's relatives?

1 485

2 484

3 468

4 467

Explanation:

A :
original image
(i) $3 W$ from husband side \& $3 M$ from wife side
$^{4} C_{3} \times^{4} C_{3} = 4 \times 4 = 16$
(ii) $3 M$ from husband side \& $3 W$ from wife side
$^{3} C_{3} \times^{3} C_{3} = 1 \times 1 = 1$
(iii) $2 W & 1 M \to$ husband
$1 W & 2 M \to$ Wife
$= (^{4} C_{2} \times^{3} C_{1}) (^{3} C_{1} \times^{4} C_{2}) = 324$
(iv) $1 W & 2 M \to$ husband
$2 W & 1 M \to$ Wife
$= (^{4} C_{1} \times^{3} C_{2}) \times (^{3} C_{2} \times^{4} C_{1})$
$= (4 \times 3) \times (3 \times 4) = 12 \times 12 = 144$ $∴$ Total $= 16 + 1 + 324 + 144 = 485$

SCRA-2009

Permutation and Combination

119011 How many numbers of five digits can be formed from the numbers $2, 0, 4, 3, 8$ , when repetition of digits is not allowed?

1 96

2 120

3 144

4 14

Explanation:

A Given, number are 2, 0, 4, 3, 8
Number of five digits numbers formed by $= 5$ !
And, number starting with zero $= 4$ !
$∴$ Required no. of five digits $= 5$ ! -4 !
$= 5 \times 4 \times 3 \times 2 - 4 \times 3 \times 2$
$= 120 - 24 = 96$

CG PET- 2005

Permutation and Combination

119013 The sides $AB, BC, CA$ of triangle $ABC$ have 3 , 4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices

1 201

2 120

3 205

4 435

Explanation:

C Given, the sides of $△ ABC$ are $AB, BC, CA$
which have 3,4 and 5 interior points.
$∴$ No. of triangles are formed using these points are $^{12} C_{3}$
For $AB$ no. of triangle forms $=^{3} C_{3}$
For BC no. of triangle forms $=^{4} C_{3}$
For $A C$ no. of triangle forms $=^{5} C_{3}$
The number of triangles which can be formed using the
points-
$=^{12} C_{3} -^{3} C_{3} -^{4} C_{3} -^{5} C_{3}$
$= \frac{12!}{9! 3!} - \frac{3!}{3! 0!} - \frac{4!}{3! 1!} - \frac{5!}{3! 2!} = 220 - 1 - 4 - 10$
$= 205$

CG PET- 2008

Permutation and Combination

119008 3 persons have 4 coats, 5 waist coats and 6 hats. The number of ways in which they put on the clothes are :

1

4^{3} \times 5^{3} \times 6^{3}

2

4 \times 5 \times 6

3

4! 5! 6

!

4 none of these

Explanation:

A : Given that,
3 person have 4 coats, 5 waist coat and 6 hats
Number of ways in which 3 men can wear 4 coats $=^{4} P_{3}$
Number of ways in which 3 men can bear 5 waistcoats
$=^{5} P_{3}$
Number of ways in which 3 men can bear 6 hats $=^{6} P_{3}$
$∴$ The total number of ways $=^{4} P_{3} \times^{5} P_{3} \times^{6} P_{3}$
$= \frac{4!}{1!} \times \frac{5!}{2!} \times \frac{6!}{3!}$
$= 24 \times 60 \times 120 = 172800$ .

BCECE-2005

Permutation and Combination

119009 A man has 7 relatives, 4 women and 3 men, His wife also has 7 relatives, 3 women and 4 men. What is the number of ways can they invite 3 women and 3 men so that 3 of them are the man's relatives and 3 of them are his wife's relatives?

1 485

2 484

3 468

4 467

Explanation:

A :
original image
(i) $3 W$ from husband side \& $3 M$ from wife side
$^{4} C_{3} \times^{4} C_{3} = 4 \times 4 = 16$
(ii) $3 M$ from husband side \& $3 W$ from wife side
$^{3} C_{3} \times^{3} C_{3} = 1 \times 1 = 1$
(iii) $2 W & 1 M \to$ husband
$1 W & 2 M \to$ Wife
$= (^{4} C_{2} \times^{3} C_{1}) (^{3} C_{1} \times^{4} C_{2}) = 324$
(iv) $1 W & 2 M \to$ husband
$2 W & 1 M \to$ Wife
$= (^{4} C_{1} \times^{3} C_{2}) \times (^{3} C_{2} \times^{4} C_{1})$
$= (4 \times 3) \times (3 \times 4) = 12 \times 12 = 144$ $∴$ Total $= 16 + 1 + 324 + 144 = 485$

SCRA-2009

Permutation and Combination

119011 How many numbers of five digits can be formed from the numbers $2, 0, 4, 3, 8$ , when repetition of digits is not allowed?

1 96

2 120

3 144

4 14

Explanation:

A Given, number are 2, 0, 4, 3, 8
Number of five digits numbers formed by $= 5$ !
And, number starting with zero $= 4$ !
$∴$ Required no. of five digits $= 5$ ! -4 !
$= 5 \times 4 \times 3 \times 2 - 4 \times 3 \times 2$
$= 120 - 24 = 96$

CG PET- 2005

Permutation and Combination

119013 The sides $AB, BC, CA$ of triangle $ABC$ have 3 , 4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices

1 201

2 120

3 205

4 435

Explanation:

C Given, the sides of $△ ABC$ are $AB, BC, CA$
which have 3,4 and 5 interior points.
$∴$ No. of triangles are formed using these points are $^{12} C_{3}$
For $AB$ no. of triangle forms $=^{3} C_{3}$
For BC no. of triangle forms $=^{4} C_{3}$
For $A C$ no. of triangle forms $=^{5} C_{3}$
The number of triangles which can be formed using the
points-
$=^{12} C_{3} -^{3} C_{3} -^{4} C_{3} -^{5} C_{3}$
$= \frac{12!}{9! 3!} - \frac{3!}{3! 0!} - \frac{4!}{3! 1!} - \frac{5!}{3! 2!} = 220 - 1 - 4 - 10$
$= 205$

CG PET- 2008

Permutation and Combination

119008 3 persons have 4 coats, 5 waist coats and 6 hats. The number of ways in which they put on the clothes are :

1

4^{3} \times 5^{3} \times 6^{3}

2

4 \times 5 \times 6

3

4! 5! 6

!

4 none of these

Explanation:

A : Given that,
3 person have 4 coats, 5 waist coat and 6 hats
Number of ways in which 3 men can wear 4 coats $=^{4} P_{3}$
Number of ways in which 3 men can bear 5 waistcoats
$=^{5} P_{3}$
Number of ways in which 3 men can bear 6 hats $=^{6} P_{3}$
$∴$ The total number of ways $=^{4} P_{3} \times^{5} P_{3} \times^{6} P_{3}$
$= \frac{4!}{1!} \times \frac{5!}{2!} \times \frac{6!}{3!}$
$= 24 \times 60 \times 120 = 172800$ .

BCECE-2005

Permutation and Combination

119009 A man has 7 relatives, 4 women and 3 men, His wife also has 7 relatives, 3 women and 4 men. What is the number of ways can they invite 3 women and 3 men so that 3 of them are the man's relatives and 3 of them are his wife's relatives?

1 485

2 484

3 468

4 467

Explanation:

A :
original image
(i) $3 W$ from husband side \& $3 M$ from wife side
$^{4} C_{3} \times^{4} C_{3} = 4 \times 4 = 16$
(ii) $3 M$ from husband side \& $3 W$ from wife side
$^{3} C_{3} \times^{3} C_{3} = 1 \times 1 = 1$
(iii) $2 W & 1 M \to$ husband
$1 W & 2 M \to$ Wife
$= (^{4} C_{2} \times^{3} C_{1}) (^{3} C_{1} \times^{4} C_{2}) = 324$
(iv) $1 W & 2 M \to$ husband
$2 W & 1 M \to$ Wife
$= (^{4} C_{1} \times^{3} C_{2}) \times (^{3} C_{2} \times^{4} C_{1})$
$= (4 \times 3) \times (3 \times 4) = 12 \times 12 = 144$ $∴$ Total $= 16 + 1 + 324 + 144 = 485$

SCRA-2009

Permutation and Combination

119011 How many numbers of five digits can be formed from the numbers $2, 0, 4, 3, 8$ , when repetition of digits is not allowed?

1 96

2 120

3 144

4 14

Explanation:

A Given, number are 2, 0, 4, 3, 8
Number of five digits numbers formed by $= 5$ !
And, number starting with zero $= 4$ !
$∴$ Required no. of five digits $= 5$ ! -4 !
$= 5 \times 4 \times 3 \times 2 - 4 \times 3 \times 2$
$= 120 - 24 = 96$

CG PET- 2005

Permutation and Combination

119013 The sides $AB, BC, CA$ of triangle $ABC$ have 3 , 4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices

1 201

2 120

3 205

4 435

Explanation:

C Given, the sides of $△ ABC$ are $AB, BC, CA$
which have 3,4 and 5 interior points.
$∴$ No. of triangles are formed using these points are $^{12} C_{3}$
For $AB$ no. of triangle forms $=^{3} C_{3}$
For BC no. of triangle forms $=^{4} C_{3}$
For $A C$ no. of triangle forms $=^{5} C_{3}$
The number of triangles which can be formed using the
points-
$=^{12} C_{3} -^{3} C_{3} -^{4} C_{3} -^{5} C_{3}$
$= \frac{12!}{9! 3!} - \frac{3!}{3! 0!} - \frac{4!}{3! 1!} - \frac{5!}{3! 2!} = 220 - 1 - 4 - 10$
$= 205$

CG PET- 2008

Permutation and Combination

119008 3 persons have 4 coats, 5 waist coats and 6 hats. The number of ways in which they put on the clothes are :

1

4^{3} \times 5^{3} \times 6^{3}

2

4 \times 5 \times 6

3

4! 5! 6

!

4 none of these

Explanation:

A : Given that,
3 person have 4 coats, 5 waist coat and 6 hats
Number of ways in which 3 men can wear 4 coats $=^{4} P_{3}$
Number of ways in which 3 men can bear 5 waistcoats
$=^{5} P_{3}$
Number of ways in which 3 men can bear 6 hats $=^{6} P_{3}$
$∴$ The total number of ways $=^{4} P_{3} \times^{5} P_{3} \times^{6} P_{3}$
$= \frac{4!}{1!} \times \frac{5!}{2!} \times \frac{6!}{3!}$
$= 24 \times 60 \times 120 = 172800$ .

BCECE-2005

Permutation and Combination

119009 A man has 7 relatives, 4 women and 3 men, His wife also has 7 relatives, 3 women and 4 men. What is the number of ways can they invite 3 women and 3 men so that 3 of them are the man's relatives and 3 of them are his wife's relatives?

1 485

2 484

3 468

4 467

Explanation:

A :
original image
(i) $3 W$ from husband side \& $3 M$ from wife side
$^{4} C_{3} \times^{4} C_{3} = 4 \times 4 = 16$
(ii) $3 M$ from husband side \& $3 W$ from wife side
$^{3} C_{3} \times^{3} C_{3} = 1 \times 1 = 1$
(iii) $2 W & 1 M \to$ husband
$1 W & 2 M \to$ Wife
$= (^{4} C_{2} \times^{3} C_{1}) (^{3} C_{1} \times^{4} C_{2}) = 324$
(iv) $1 W & 2 M \to$ husband
$2 W & 1 M \to$ Wife
$= (^{4} C_{1} \times^{3} C_{2}) \times (^{3} C_{2} \times^{4} C_{1})$
$= (4 \times 3) \times (3 \times 4) = 12 \times 12 = 144$ $∴$ Total $= 16 + 1 + 324 + 144 = 485$

SCRA-2009

Permutation and Combination

119011 How many numbers of five digits can be formed from the numbers $2, 0, 4, 3, 8$ , when repetition of digits is not allowed?

1 96

2 120

3 144

4 14

Explanation:

A Given, number are 2, 0, 4, 3, 8
Number of five digits numbers formed by $= 5$ !
And, number starting with zero $= 4$ !
$∴$ Required no. of five digits $= 5$ ! -4 !
$= 5 \times 4 \times 3 \times 2 - 4 \times 3 \times 2$
$= 120 - 24 = 96$

CG PET- 2005

Permutation and Combination

119013 The sides $AB, BC, CA$ of triangle $ABC$ have 3 , 4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices

1 201

2 120

3 205

4 435

Explanation:

C Given, the sides of $△ ABC$ are $AB, BC, CA$
which have 3,4 and 5 interior points.
$∴$ No. of triangles are formed using these points are $^{12} C_{3}$
For $AB$ no. of triangle forms $=^{3} C_{3}$
For BC no. of triangle forms $=^{4} C_{3}$
For $A C$ no. of triangle forms $=^{5} C_{3}$
The number of triangles which can be formed using the
points-
$=^{12} C_{3} -^{3} C_{3} -^{4} C_{3} -^{5} C_{3}$
$= \frac{12!}{9! 3!} - \frac{3!}{3! 0!} - \frac{4!}{3! 1!} - \frac{5!}{3! 2!} = 220 - 1 - 4 - 10$
$= 205$

CG PET- 2008