NEET Test Series from KOTA - 10 Papers In MS WORD
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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 \(={ }^4 \mathrm{P}_3\) Number of ways in which 3 men can bear 5 waistcoats \(={ }^5 \mathrm{P}_3\) Number of ways in which 3 men can bear 6 hats \(={ }^6 \mathrm{P}_3\) \(\therefore\) The total number of ways \(={ }^4 \mathrm{P}_3 \times{ }^5 \mathrm{P}_3 \times{ }^6 \mathrm{P}_3\) \(=\frac{4 !}{1 !} \times \frac{5 !}{2 !} \times \frac{6 !}{3 !}\) \(=24 \times 60 \times 120=172800\).
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 : (i) \(3 \mathrm{~W}\) from husband side \& \(3 \mathrm{M}\) from wife side \({ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3=4 \times 4=16\) (ii) \(3 \mathrm{M}\) from husband side \& \(3 \mathrm{~W}\) from wife side \({ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_3=1 \times 1=1\) (iii) \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) husband \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1\right)\left({ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2\right)=324\) (iv) \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) husband \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2\right) \times\left({ }^3 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1\right)\) \(=(4 \times 3) \times(3 \times 4)=12 \times 12=144\)\(\therefore\) 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 \(=4\) ! \(\therefore\) 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 \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) of triangle \(\mathrm{ABC}\) 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 \(\triangle \mathrm{ABC}\) are \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) which have 3,4 and 5 interior points. \(\therefore\) No. of triangles are formed using these points are \({ }^{12} \mathrm{C}_3\) For \(\mathrm{AB}\) no. of triangle forms \(={ }^3 \mathrm{C}_3\) For BC no. of triangle forms \(={ }^4 \mathrm{C}_3\) For \(A C\) no. of triangle forms \(={ }^5 \mathrm{C}_3\) The number of triangles which can be formed using the points- \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3\) \(=\frac{12 !}{9 ! 3 !}-\frac{3 !}{3 ! 0 !}-\frac{4 !}{3 ! 1 !}-\frac{5 !}{3 ! 2 !}=220-1-4-10\) \(=205\)
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 \(={ }^4 \mathrm{P}_3\) Number of ways in which 3 men can bear 5 waistcoats \(={ }^5 \mathrm{P}_3\) Number of ways in which 3 men can bear 6 hats \(={ }^6 \mathrm{P}_3\) \(\therefore\) The total number of ways \(={ }^4 \mathrm{P}_3 \times{ }^5 \mathrm{P}_3 \times{ }^6 \mathrm{P}_3\) \(=\frac{4 !}{1 !} \times \frac{5 !}{2 !} \times \frac{6 !}{3 !}\) \(=24 \times 60 \times 120=172800\).
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 : (i) \(3 \mathrm{~W}\) from husband side \& \(3 \mathrm{M}\) from wife side \({ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3=4 \times 4=16\) (ii) \(3 \mathrm{M}\) from husband side \& \(3 \mathrm{~W}\) from wife side \({ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_3=1 \times 1=1\) (iii) \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) husband \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1\right)\left({ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2\right)=324\) (iv) \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) husband \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2\right) \times\left({ }^3 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1\right)\) \(=(4 \times 3) \times(3 \times 4)=12 \times 12=144\)\(\therefore\) 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 \(=4\) ! \(\therefore\) 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 \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) of triangle \(\mathrm{ABC}\) 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 \(\triangle \mathrm{ABC}\) are \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) which have 3,4 and 5 interior points. \(\therefore\) No. of triangles are formed using these points are \({ }^{12} \mathrm{C}_3\) For \(\mathrm{AB}\) no. of triangle forms \(={ }^3 \mathrm{C}_3\) For BC no. of triangle forms \(={ }^4 \mathrm{C}_3\) For \(A C\) no. of triangle forms \(={ }^5 \mathrm{C}_3\) The number of triangles which can be formed using the points- \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3\) \(=\frac{12 !}{9 ! 3 !}-\frac{3 !}{3 ! 0 !}-\frac{4 !}{3 ! 1 !}-\frac{5 !}{3 ! 2 !}=220-1-4-10\) \(=205\)
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 \(={ }^4 \mathrm{P}_3\) Number of ways in which 3 men can bear 5 waistcoats \(={ }^5 \mathrm{P}_3\) Number of ways in which 3 men can bear 6 hats \(={ }^6 \mathrm{P}_3\) \(\therefore\) The total number of ways \(={ }^4 \mathrm{P}_3 \times{ }^5 \mathrm{P}_3 \times{ }^6 \mathrm{P}_3\) \(=\frac{4 !}{1 !} \times \frac{5 !}{2 !} \times \frac{6 !}{3 !}\) \(=24 \times 60 \times 120=172800\).
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 : (i) \(3 \mathrm{~W}\) from husband side \& \(3 \mathrm{M}\) from wife side \({ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3=4 \times 4=16\) (ii) \(3 \mathrm{M}\) from husband side \& \(3 \mathrm{~W}\) from wife side \({ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_3=1 \times 1=1\) (iii) \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) husband \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1\right)\left({ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2\right)=324\) (iv) \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) husband \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2\right) \times\left({ }^3 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1\right)\) \(=(4 \times 3) \times(3 \times 4)=12 \times 12=144\)\(\therefore\) 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 \(=4\) ! \(\therefore\) 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 \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) of triangle \(\mathrm{ABC}\) 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 \(\triangle \mathrm{ABC}\) are \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) which have 3,4 and 5 interior points. \(\therefore\) No. of triangles are formed using these points are \({ }^{12} \mathrm{C}_3\) For \(\mathrm{AB}\) no. of triangle forms \(={ }^3 \mathrm{C}_3\) For BC no. of triangle forms \(={ }^4 \mathrm{C}_3\) For \(A C\) no. of triangle forms \(={ }^5 \mathrm{C}_3\) The number of triangles which can be formed using the points- \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3\) \(=\frac{12 !}{9 ! 3 !}-\frac{3 !}{3 ! 0 !}-\frac{4 !}{3 ! 1 !}-\frac{5 !}{3 ! 2 !}=220-1-4-10\) \(=205\)
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 \(={ }^4 \mathrm{P}_3\) Number of ways in which 3 men can bear 5 waistcoats \(={ }^5 \mathrm{P}_3\) Number of ways in which 3 men can bear 6 hats \(={ }^6 \mathrm{P}_3\) \(\therefore\) The total number of ways \(={ }^4 \mathrm{P}_3 \times{ }^5 \mathrm{P}_3 \times{ }^6 \mathrm{P}_3\) \(=\frac{4 !}{1 !} \times \frac{5 !}{2 !} \times \frac{6 !}{3 !}\) \(=24 \times 60 \times 120=172800\).
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 : (i) \(3 \mathrm{~W}\) from husband side \& \(3 \mathrm{M}\) from wife side \({ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3=4 \times 4=16\) (ii) \(3 \mathrm{M}\) from husband side \& \(3 \mathrm{~W}\) from wife side \({ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_3=1 \times 1=1\) (iii) \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) husband \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1\right)\left({ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2\right)=324\) (iv) \(1 \mathrm{~W} \& 2 \mathrm{M} \rightarrow\) husband \(2 \mathrm{~W} \& 1 \mathrm{M} \rightarrow\) Wife \(=\left({ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2\right) \times\left({ }^3 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1\right)\) \(=(4 \times 3) \times(3 \times 4)=12 \times 12=144\)\(\therefore\) 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 \(=4\) ! \(\therefore\) 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 \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) of triangle \(\mathrm{ABC}\) 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 \(\triangle \mathrm{ABC}\) are \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) which have 3,4 and 5 interior points. \(\therefore\) No. of triangles are formed using these points are \({ }^{12} \mathrm{C}_3\) For \(\mathrm{AB}\) no. of triangle forms \(={ }^3 \mathrm{C}_3\) For BC no. of triangle forms \(={ }^4 \mathrm{C}_3\) For \(A C\) no. of triangle forms \(={ }^5 \mathrm{C}_3\) The number of triangles which can be formed using the points- \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3\) \(=\frac{12 !}{9 ! 3 !}-\frac{3 !}{3 ! 0 !}-\frac{4 !}{3 ! 1 !}-\frac{5 !}{3 ! 2 !}=220-1-4-10\) \(=205\)
