118990
If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
1 324
2 341
3 359
4 None of these
Explanation:
A Here the number of words starting from \(\mathrm{A}\) are \(5 !=5 \times 4 \times 3 \times 2 \times 1=120\) The number of words starting from I are \(5 !=120\) The number of words starting from KA are \(4 !=4 \times 3 \times 2 \times 1=24\) The number of words starting from KI are \(4 !=24\) The number of words starting from \(\mathrm{KN}\) are \(4 !=24\) The number of words starting from KRA are 3 ! \(=3 \times 2 \times 1=6\) The number of words starting from KRIA are 2 ! \(=2 \times 1=2\) The number of words starting from KRIN are \(2 !=2\) The number of words starting from KRISA are \(1 !=1\) The number of words starting from KRISNA are \(1 !=1\) Hence, rank of word 'KRISNA' \(=2(120)+3(24)+6+2(2)+2(1)\) \(=240+72+6+4+2=324\)
VITEEE-2016
Permutation and Combination
118991
Number of ways of selecting three squares on a chessboard so that all the three be on a diagonal line of the board or parallel to it is
1 196
2 126
3 252
4 392
Explanation:
D We know that, there are total 64 squares in chess board So, number of ways \(=\left[\left({ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3+{ }^5 \mathrm{C}_3+{ }^6 \mathrm{C}_3+{ }^7 \mathrm{C}_3\right) \times 2+{ }^8 \mathrm{C}_3\right] \times 2\) \(=[(1+4+10+20+35) \times 2+56] \times 2\) \(=[(70) \times 2+56] \times 2=[140+56] \times 2\) \(=[196] \times 2=392\)
VITEEE-2016
Permutation and Combination
118992
The sides \(\mathrm{AB}, \mathrm{BC}\) and \(\mathrm{CA}\) of a \(\triangle \mathrm{ABC}\) have respectively 3,4 and 5 points lying on them. The number of triangles that can be constructed using these points as vertices is
1 205
2 220
3 210
4 None of these
Explanation:
A Given, There are 12 points out of which 3,4 and 5 are collinear. To form a triangle 3 non-collinear points are required. Total number of ways of choosing 3 points from 12 points \(={ }^{12} \mathrm{C}_3\) Now, we need to subtract 3 points which are collinear \(\therefore\) Total number of triangles that can be formed \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3=205\)
VITEEE-2013
Permutation and Combination
118993
Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?
1 120
2 80
3 90
4 100
Explanation:
D Given that, Number of boys \(=6\) Number of girls \(=4\) Making group of 7 members So, the total number of ways or combinations \(={ }^6 \mathrm{C}_4 \times{ }^4 \mathrm{C}_3+{ }^6 \mathrm{C}_5 \times{ }^4 \mathrm{C}_2+{ }^6 \mathrm{C}_6 \times{ }^4 \mathrm{C}_1\) \(=\frac{6 !}{4 ! 2 !} \times \frac{4 !}{3 ! 1 !}+\frac{6 !}{5 ! 1 !} \times \frac{4 !}{2 ! 2 !}+\frac{6 !}{6 ! 0 !} \times \frac{4 !}{1 ! 3 !}\) \(=15 \times 4+6 \times 6+1 \times 4\) \(=60+36+4=100\)
118990
If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
1 324
2 341
3 359
4 None of these
Explanation:
A Here the number of words starting from \(\mathrm{A}\) are \(5 !=5 \times 4 \times 3 \times 2 \times 1=120\) The number of words starting from I are \(5 !=120\) The number of words starting from KA are \(4 !=4 \times 3 \times 2 \times 1=24\) The number of words starting from KI are \(4 !=24\) The number of words starting from \(\mathrm{KN}\) are \(4 !=24\) The number of words starting from KRA are 3 ! \(=3 \times 2 \times 1=6\) The number of words starting from KRIA are 2 ! \(=2 \times 1=2\) The number of words starting from KRIN are \(2 !=2\) The number of words starting from KRISA are \(1 !=1\) The number of words starting from KRISNA are \(1 !=1\) Hence, rank of word 'KRISNA' \(=2(120)+3(24)+6+2(2)+2(1)\) \(=240+72+6+4+2=324\)
VITEEE-2016
Permutation and Combination
118991
Number of ways of selecting three squares on a chessboard so that all the three be on a diagonal line of the board or parallel to it is
1 196
2 126
3 252
4 392
Explanation:
D We know that, there are total 64 squares in chess board So, number of ways \(=\left[\left({ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3+{ }^5 \mathrm{C}_3+{ }^6 \mathrm{C}_3+{ }^7 \mathrm{C}_3\right) \times 2+{ }^8 \mathrm{C}_3\right] \times 2\) \(=[(1+4+10+20+35) \times 2+56] \times 2\) \(=[(70) \times 2+56] \times 2=[140+56] \times 2\) \(=[196] \times 2=392\)
VITEEE-2016
Permutation and Combination
118992
The sides \(\mathrm{AB}, \mathrm{BC}\) and \(\mathrm{CA}\) of a \(\triangle \mathrm{ABC}\) have respectively 3,4 and 5 points lying on them. The number of triangles that can be constructed using these points as vertices is
1 205
2 220
3 210
4 None of these
Explanation:
A Given, There are 12 points out of which 3,4 and 5 are collinear. To form a triangle 3 non-collinear points are required. Total number of ways of choosing 3 points from 12 points \(={ }^{12} \mathrm{C}_3\) Now, we need to subtract 3 points which are collinear \(\therefore\) Total number of triangles that can be formed \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3=205\)
VITEEE-2013
Permutation and Combination
118993
Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?
1 120
2 80
3 90
4 100
Explanation:
D Given that, Number of boys \(=6\) Number of girls \(=4\) Making group of 7 members So, the total number of ways or combinations \(={ }^6 \mathrm{C}_4 \times{ }^4 \mathrm{C}_3+{ }^6 \mathrm{C}_5 \times{ }^4 \mathrm{C}_2+{ }^6 \mathrm{C}_6 \times{ }^4 \mathrm{C}_1\) \(=\frac{6 !}{4 ! 2 !} \times \frac{4 !}{3 ! 1 !}+\frac{6 !}{5 ! 1 !} \times \frac{4 !}{2 ! 2 !}+\frac{6 !}{6 ! 0 !} \times \frac{4 !}{1 ! 3 !}\) \(=15 \times 4+6 \times 6+1 \times 4\) \(=60+36+4=100\)
118990
If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
1 324
2 341
3 359
4 None of these
Explanation:
A Here the number of words starting from \(\mathrm{A}\) are \(5 !=5 \times 4 \times 3 \times 2 \times 1=120\) The number of words starting from I are \(5 !=120\) The number of words starting from KA are \(4 !=4 \times 3 \times 2 \times 1=24\) The number of words starting from KI are \(4 !=24\) The number of words starting from \(\mathrm{KN}\) are \(4 !=24\) The number of words starting from KRA are 3 ! \(=3 \times 2 \times 1=6\) The number of words starting from KRIA are 2 ! \(=2 \times 1=2\) The number of words starting from KRIN are \(2 !=2\) The number of words starting from KRISA are \(1 !=1\) The number of words starting from KRISNA are \(1 !=1\) Hence, rank of word 'KRISNA' \(=2(120)+3(24)+6+2(2)+2(1)\) \(=240+72+6+4+2=324\)
VITEEE-2016
Permutation and Combination
118991
Number of ways of selecting three squares on a chessboard so that all the three be on a diagonal line of the board or parallel to it is
1 196
2 126
3 252
4 392
Explanation:
D We know that, there are total 64 squares in chess board So, number of ways \(=\left[\left({ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3+{ }^5 \mathrm{C}_3+{ }^6 \mathrm{C}_3+{ }^7 \mathrm{C}_3\right) \times 2+{ }^8 \mathrm{C}_3\right] \times 2\) \(=[(1+4+10+20+35) \times 2+56] \times 2\) \(=[(70) \times 2+56] \times 2=[140+56] \times 2\) \(=[196] \times 2=392\)
VITEEE-2016
Permutation and Combination
118992
The sides \(\mathrm{AB}, \mathrm{BC}\) and \(\mathrm{CA}\) of a \(\triangle \mathrm{ABC}\) have respectively 3,4 and 5 points lying on them. The number of triangles that can be constructed using these points as vertices is
1 205
2 220
3 210
4 None of these
Explanation:
A Given, There are 12 points out of which 3,4 and 5 are collinear. To form a triangle 3 non-collinear points are required. Total number of ways of choosing 3 points from 12 points \(={ }^{12} \mathrm{C}_3\) Now, we need to subtract 3 points which are collinear \(\therefore\) Total number of triangles that can be formed \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3=205\)
VITEEE-2013
Permutation and Combination
118993
Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?
1 120
2 80
3 90
4 100
Explanation:
D Given that, Number of boys \(=6\) Number of girls \(=4\) Making group of 7 members So, the total number of ways or combinations \(={ }^6 \mathrm{C}_4 \times{ }^4 \mathrm{C}_3+{ }^6 \mathrm{C}_5 \times{ }^4 \mathrm{C}_2+{ }^6 \mathrm{C}_6 \times{ }^4 \mathrm{C}_1\) \(=\frac{6 !}{4 ! 2 !} \times \frac{4 !}{3 ! 1 !}+\frac{6 !}{5 ! 1 !} \times \frac{4 !}{2 ! 2 !}+\frac{6 !}{6 ! 0 !} \times \frac{4 !}{1 ! 3 !}\) \(=15 \times 4+6 \times 6+1 \times 4\) \(=60+36+4=100\)
118990
If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
1 324
2 341
3 359
4 None of these
Explanation:
A Here the number of words starting from \(\mathrm{A}\) are \(5 !=5 \times 4 \times 3 \times 2 \times 1=120\) The number of words starting from I are \(5 !=120\) The number of words starting from KA are \(4 !=4 \times 3 \times 2 \times 1=24\) The number of words starting from KI are \(4 !=24\) The number of words starting from \(\mathrm{KN}\) are \(4 !=24\) The number of words starting from KRA are 3 ! \(=3 \times 2 \times 1=6\) The number of words starting from KRIA are 2 ! \(=2 \times 1=2\) The number of words starting from KRIN are \(2 !=2\) The number of words starting from KRISA are \(1 !=1\) The number of words starting from KRISNA are \(1 !=1\) Hence, rank of word 'KRISNA' \(=2(120)+3(24)+6+2(2)+2(1)\) \(=240+72+6+4+2=324\)
VITEEE-2016
Permutation and Combination
118991
Number of ways of selecting three squares on a chessboard so that all the three be on a diagonal line of the board or parallel to it is
1 196
2 126
3 252
4 392
Explanation:
D We know that, there are total 64 squares in chess board So, number of ways \(=\left[\left({ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3+{ }^5 \mathrm{C}_3+{ }^6 \mathrm{C}_3+{ }^7 \mathrm{C}_3\right) \times 2+{ }^8 \mathrm{C}_3\right] \times 2\) \(=[(1+4+10+20+35) \times 2+56] \times 2\) \(=[(70) \times 2+56] \times 2=[140+56] \times 2\) \(=[196] \times 2=392\)
VITEEE-2016
Permutation and Combination
118992
The sides \(\mathrm{AB}, \mathrm{BC}\) and \(\mathrm{CA}\) of a \(\triangle \mathrm{ABC}\) have respectively 3,4 and 5 points lying on them. The number of triangles that can be constructed using these points as vertices is
1 205
2 220
3 210
4 None of these
Explanation:
A Given, There are 12 points out of which 3,4 and 5 are collinear. To form a triangle 3 non-collinear points are required. Total number of ways of choosing 3 points from 12 points \(={ }^{12} \mathrm{C}_3\) Now, we need to subtract 3 points which are collinear \(\therefore\) Total number of triangles that can be formed \(={ }^{12} \mathrm{C}_3-{ }^3 \mathrm{C}_3-{ }^4 \mathrm{C}_3-{ }^5 \mathrm{C}_3=205\)
VITEEE-2013
Permutation and Combination
118993
Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?
1 120
2 80
3 90
4 100
Explanation:
D Given that, Number of boys \(=6\) Number of girls \(=4\) Making group of 7 members So, the total number of ways or combinations \(={ }^6 \mathrm{C}_4 \times{ }^4 \mathrm{C}_3+{ }^6 \mathrm{C}_5 \times{ }^4 \mathrm{C}_2+{ }^6 \mathrm{C}_6 \times{ }^4 \mathrm{C}_1\) \(=\frac{6 !}{4 ! 2 !} \times \frac{4 !}{3 ! 1 !}+\frac{6 !}{5 ! 1 !} \times \frac{4 !}{2 ! 2 !}+\frac{6 !}{6 ! 0 !} \times \frac{4 !}{1 ! 3 !}\) \(=15 \times 4+6 \times 6+1 \times 4\) \(=60+36+4=100\)
