119286
A three digit number \(n\) is such that the last two digits of it are equal and differ from the first. The number of \(n\) 's is
1 64
2 72
3 81
4 900
Explanation:
C Here hundreds place, can be filled with digits 1 to 9 (except 0 ). tens place can be filled in 9 ways (all digits except the one at hundreds place) Now the unit place can be filled in only one way since last two digit are equal and differ from first \(\therefore\) Total number of ways to write such a number \(=9 \times 9 \times 1\) \(=81 \times 1\) \(=81\)
AP EAMCET-2005
Permutation and Combination
119287
The digit in the unit's place of the number 1 ! + \(2 !+3 !+\ldots \ldots+99\) ! is
1 3
2 0
3 1
4 7
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots+99 !\) we have to evaluate unit's digit of the sum. \(1 !+2 !+3 !+4 !+5 !+6 ! \ldots . .+99 !\) As we know that \(1 !=1\) \(2 !=2 \times 1=2 \quad 5 !=5 \times 4 \times 3 \times 2 \times 1=120\) \(3 !=3 \times 2 \times 1=6 \quad 7 !=7 \times 5 \times 4 \times 3 \times 2 \times 1=720\) \(4 !=4 \times 3 \times 2 \times 1=24\) Clearly factorial of every no. after 4 will multiple of 10 So, factorial value every no. after 4 will have 0 at its unit place. Therefore, unit digit of required sum \(=1+2+6+4+0\) \(=13\) Hence the required unit digit is 3 .
WB JEE-2021
Permutation and Combination
119288
The number of zeros at the end of 100 is
1 21
2 22
3 23
4 24
Explanation:
D The power of 5 is the sum of \(\frac{100}{5}=20\) \(\frac{20}{5}=4\) The power of \(5=20+4\) \(=24\)
WB JEE-2022
Permutation and Combination
119289
If four points are taken on each of three parallel lines in a plane, then the maximum number of triangles formed with these points is
1 64
2 144
3 208
4 80
Explanation:
C Here, number of points \(=4\) \(\therefore\) The required number of triangle \(=\mathrm{p}^3+3 \mathrm{p}^2(\mathrm{p}-1)\) \(=4^3+3(4)^2(4-1)\) \(=64+3 \times 16 \times 3\) \(=64+144\) \(=208\)
119286
A three digit number \(n\) is such that the last two digits of it are equal and differ from the first. The number of \(n\) 's is
1 64
2 72
3 81
4 900
Explanation:
C Here hundreds place, can be filled with digits 1 to 9 (except 0 ). tens place can be filled in 9 ways (all digits except the one at hundreds place) Now the unit place can be filled in only one way since last two digit are equal and differ from first \(\therefore\) Total number of ways to write such a number \(=9 \times 9 \times 1\) \(=81 \times 1\) \(=81\)
AP EAMCET-2005
Permutation and Combination
119287
The digit in the unit's place of the number 1 ! + \(2 !+3 !+\ldots \ldots+99\) ! is
1 3
2 0
3 1
4 7
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots+99 !\) we have to evaluate unit's digit of the sum. \(1 !+2 !+3 !+4 !+5 !+6 ! \ldots . .+99 !\) As we know that \(1 !=1\) \(2 !=2 \times 1=2 \quad 5 !=5 \times 4 \times 3 \times 2 \times 1=120\) \(3 !=3 \times 2 \times 1=6 \quad 7 !=7 \times 5 \times 4 \times 3 \times 2 \times 1=720\) \(4 !=4 \times 3 \times 2 \times 1=24\) Clearly factorial of every no. after 4 will multiple of 10 So, factorial value every no. after 4 will have 0 at its unit place. Therefore, unit digit of required sum \(=1+2+6+4+0\) \(=13\) Hence the required unit digit is 3 .
WB JEE-2021
Permutation and Combination
119288
The number of zeros at the end of 100 is
1 21
2 22
3 23
4 24
Explanation:
D The power of 5 is the sum of \(\frac{100}{5}=20\) \(\frac{20}{5}=4\) The power of \(5=20+4\) \(=24\)
WB JEE-2022
Permutation and Combination
119289
If four points are taken on each of three parallel lines in a plane, then the maximum number of triangles formed with these points is
1 64
2 144
3 208
4 80
Explanation:
C Here, number of points \(=4\) \(\therefore\) The required number of triangle \(=\mathrm{p}^3+3 \mathrm{p}^2(\mathrm{p}-1)\) \(=4^3+3(4)^2(4-1)\) \(=64+3 \times 16 \times 3\) \(=64+144\) \(=208\)
119286
A three digit number \(n\) is such that the last two digits of it are equal and differ from the first. The number of \(n\) 's is
1 64
2 72
3 81
4 900
Explanation:
C Here hundreds place, can be filled with digits 1 to 9 (except 0 ). tens place can be filled in 9 ways (all digits except the one at hundreds place) Now the unit place can be filled in only one way since last two digit are equal and differ from first \(\therefore\) Total number of ways to write such a number \(=9 \times 9 \times 1\) \(=81 \times 1\) \(=81\)
AP EAMCET-2005
Permutation and Combination
119287
The digit in the unit's place of the number 1 ! + \(2 !+3 !+\ldots \ldots+99\) ! is
1 3
2 0
3 1
4 7
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots+99 !\) we have to evaluate unit's digit of the sum. \(1 !+2 !+3 !+4 !+5 !+6 ! \ldots . .+99 !\) As we know that \(1 !=1\) \(2 !=2 \times 1=2 \quad 5 !=5 \times 4 \times 3 \times 2 \times 1=120\) \(3 !=3 \times 2 \times 1=6 \quad 7 !=7 \times 5 \times 4 \times 3 \times 2 \times 1=720\) \(4 !=4 \times 3 \times 2 \times 1=24\) Clearly factorial of every no. after 4 will multiple of 10 So, factorial value every no. after 4 will have 0 at its unit place. Therefore, unit digit of required sum \(=1+2+6+4+0\) \(=13\) Hence the required unit digit is 3 .
WB JEE-2021
Permutation and Combination
119288
The number of zeros at the end of 100 is
1 21
2 22
3 23
4 24
Explanation:
D The power of 5 is the sum of \(\frac{100}{5}=20\) \(\frac{20}{5}=4\) The power of \(5=20+4\) \(=24\)
WB JEE-2022
Permutation and Combination
119289
If four points are taken on each of three parallel lines in a plane, then the maximum number of triangles formed with these points is
1 64
2 144
3 208
4 80
Explanation:
C Here, number of points \(=4\) \(\therefore\) The required number of triangle \(=\mathrm{p}^3+3 \mathrm{p}^2(\mathrm{p}-1)\) \(=4^3+3(4)^2(4-1)\) \(=64+3 \times 16 \times 3\) \(=64+144\) \(=208\)
119286
A three digit number \(n\) is such that the last two digits of it are equal and differ from the first. The number of \(n\) 's is
1 64
2 72
3 81
4 900
Explanation:
C Here hundreds place, can be filled with digits 1 to 9 (except 0 ). tens place can be filled in 9 ways (all digits except the one at hundreds place) Now the unit place can be filled in only one way since last two digit are equal and differ from first \(\therefore\) Total number of ways to write such a number \(=9 \times 9 \times 1\) \(=81 \times 1\) \(=81\)
AP EAMCET-2005
Permutation and Combination
119287
The digit in the unit's place of the number 1 ! + \(2 !+3 !+\ldots \ldots+99\) ! is
1 3
2 0
3 1
4 7
Explanation:
A Given that, \(1 !+2 !+3 !+\ldots+99 !\) we have to evaluate unit's digit of the sum. \(1 !+2 !+3 !+4 !+5 !+6 ! \ldots . .+99 !\) As we know that \(1 !=1\) \(2 !=2 \times 1=2 \quad 5 !=5 \times 4 \times 3 \times 2 \times 1=120\) \(3 !=3 \times 2 \times 1=6 \quad 7 !=7 \times 5 \times 4 \times 3 \times 2 \times 1=720\) \(4 !=4 \times 3 \times 2 \times 1=24\) Clearly factorial of every no. after 4 will multiple of 10 So, factorial value every no. after 4 will have 0 at its unit place. Therefore, unit digit of required sum \(=1+2+6+4+0\) \(=13\) Hence the required unit digit is 3 .
WB JEE-2021
Permutation and Combination
119288
The number of zeros at the end of 100 is
1 21
2 22
3 23
4 24
Explanation:
D The power of 5 is the sum of \(\frac{100}{5}=20\) \(\frac{20}{5}=4\) The power of \(5=20+4\) \(=24\)
WB JEE-2022
Permutation and Combination
119289
If four points are taken on each of three parallel lines in a plane, then the maximum number of triangles formed with these points is
1 64
2 144
3 208
4 80
Explanation:
C Here, number of points \(=4\) \(\therefore\) The required number of triangle \(=\mathrm{p}^3+3 \mathrm{p}^2(\mathrm{p}-1)\) \(=4^3+3(4)^2(4-1)\) \(=64+3 \times 16 \times 3\) \(=64+144\) \(=208\)
