NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
119297
Statements I The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty, is \({ }^9 \mathrm{C}_3\). Statements II The number of ways of choosing any 3 place from 9 different places is \({ }^9 \mathrm{C}_3\).
1 Statement I is true, Statement II is true; Statements II is not a correct explanation of Statement I
2 Statement I is true, Statement II is false
3 Statement I is false, Statement II is true
4 Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
Explanation:
A Statement (I) Given, \(\mathrm{n}=10, \mathrm{r}=4\) Required no of ways \(={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{10-1} \mathrm{C}_{4-1}\) \(={ }^9 \mathrm{C}_3\) Statement (II) there are 9 different places. Required no of ways \(={ }^9 \mathrm{C}_3\) Hence both statement are correct but \(\mathrm{II}^{\text {nd }}\) statement is not a correct explanation of \(\mathrm{I}^{\text {st }}\) statement.
AIEEE-2011
Permutation and Combination
119298
There are 10 points in a plane, out of these 6 are collinear. If \(\mathbf{N}\) is the number of triangles formed by joining these points, then
1 \(\mathrm{N}>190\)
2 \(\mathrm{N} \leq 100\)
3 \(100\lt \mathrm{N} \leq 140\)
4 \(140\lt \mathrm{N} \leq 190\)
Explanation:
B Given, \(\mathrm{n}=10\) points, \(\mathrm{m}=6\) If out of \(n\) points \(m\) are collinear then no of triangles \(={ }^{\mathrm{n}} \mathrm{C}_3-{ }^{\mathrm{m}} \mathrm{C}_3\) \(={ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=\frac{10 !}{3 !(10-3) !}-\frac{6 !}{3 !(6-3) !}\) \(=\frac{10.8 .9 .7 !}{3.2 .17 !}-\frac{6.5 .4 .3 \text { ! }}{3.23 !}\) \(=120-20\) \(=100\)
AIEEE-2011
Permutation and Combination
119300
For which value of \(n \in N, n, i\) has 13 trailing zeros?
1 51
2 54
3 57
4 60
Explanation:
C \(n\) ! exponent of 5 is always less than of equal to exponent of 2 \(13=\frac{\mathrm{n}}{5}+\frac{\mathrm{n}}{5^2}+\frac{\mathrm{n}}{5^3}+\ldots \ldots \ldots .\) According to option \(\mathrm{n}=57\) \(\frac{57}{5}+\frac{57}{25}+\frac{57}{125}\) \(=11+2+\ldots \ldots 0\)Hence in 57 ! the power of 5 is 13
AP EAMCET-23.09.2020
Permutation and Combination
119301
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choice available to him is
1 196
2 140
3 168
4 176
Explanation:
A If we choose 4 then no of choice to choose 4 from 5 is \({ }^5 \mathrm{C}_4\) \({ }^5 \mathrm{C}_4 \times{ }^8 \mathrm{C}_6\) \(\frac{5 !}{4 !(5-4) !} \times \frac{8 !}{6 !(8-6) !}\) \(\frac{5 !}{4 ! 1 !} \times \frac{8.76 . !}{6.1 .2 !}\) \(5 \times 28=140\) Selecting five out of first five question and five out of remaining eight questions \({ }^5 \mathrm{C}_5 \times{ }^8 \mathrm{C}_5\) \(=1 \times \frac{8 !}{5 !(8-5) !}\) \(=1 \times \frac{8 \times 7 \times 6 \times 5 !}{5 ! 3 !}\) \(=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}\) \(=56\)Hence, the total no of choice \(140+96=196\)
119297
Statements I The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty, is \({ }^9 \mathrm{C}_3\). Statements II The number of ways of choosing any 3 place from 9 different places is \({ }^9 \mathrm{C}_3\).
1 Statement I is true, Statement II is true; Statements II is not a correct explanation of Statement I
2 Statement I is true, Statement II is false
3 Statement I is false, Statement II is true
4 Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
Explanation:
A Statement (I) Given, \(\mathrm{n}=10, \mathrm{r}=4\) Required no of ways \(={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{10-1} \mathrm{C}_{4-1}\) \(={ }^9 \mathrm{C}_3\) Statement (II) there are 9 different places. Required no of ways \(={ }^9 \mathrm{C}_3\) Hence both statement are correct but \(\mathrm{II}^{\text {nd }}\) statement is not a correct explanation of \(\mathrm{I}^{\text {st }}\) statement.
AIEEE-2011
Permutation and Combination
119298
There are 10 points in a plane, out of these 6 are collinear. If \(\mathbf{N}\) is the number of triangles formed by joining these points, then
1 \(\mathrm{N}>190\)
2 \(\mathrm{N} \leq 100\)
3 \(100\lt \mathrm{N} \leq 140\)
4 \(140\lt \mathrm{N} \leq 190\)
Explanation:
B Given, \(\mathrm{n}=10\) points, \(\mathrm{m}=6\) If out of \(n\) points \(m\) are collinear then no of triangles \(={ }^{\mathrm{n}} \mathrm{C}_3-{ }^{\mathrm{m}} \mathrm{C}_3\) \(={ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=\frac{10 !}{3 !(10-3) !}-\frac{6 !}{3 !(6-3) !}\) \(=\frac{10.8 .9 .7 !}{3.2 .17 !}-\frac{6.5 .4 .3 \text { ! }}{3.23 !}\) \(=120-20\) \(=100\)
AIEEE-2011
Permutation and Combination
119300
For which value of \(n \in N, n, i\) has 13 trailing zeros?
1 51
2 54
3 57
4 60
Explanation:
C \(n\) ! exponent of 5 is always less than of equal to exponent of 2 \(13=\frac{\mathrm{n}}{5}+\frac{\mathrm{n}}{5^2}+\frac{\mathrm{n}}{5^3}+\ldots \ldots \ldots .\) According to option \(\mathrm{n}=57\) \(\frac{57}{5}+\frac{57}{25}+\frac{57}{125}\) \(=11+2+\ldots \ldots 0\)Hence in 57 ! the power of 5 is 13
AP EAMCET-23.09.2020
Permutation and Combination
119301
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choice available to him is
1 196
2 140
3 168
4 176
Explanation:
A If we choose 4 then no of choice to choose 4 from 5 is \({ }^5 \mathrm{C}_4\) \({ }^5 \mathrm{C}_4 \times{ }^8 \mathrm{C}_6\) \(\frac{5 !}{4 !(5-4) !} \times \frac{8 !}{6 !(8-6) !}\) \(\frac{5 !}{4 ! 1 !} \times \frac{8.76 . !}{6.1 .2 !}\) \(5 \times 28=140\) Selecting five out of first five question and five out of remaining eight questions \({ }^5 \mathrm{C}_5 \times{ }^8 \mathrm{C}_5\) \(=1 \times \frac{8 !}{5 !(8-5) !}\) \(=1 \times \frac{8 \times 7 \times 6 \times 5 !}{5 ! 3 !}\) \(=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}\) \(=56\)Hence, the total no of choice \(140+96=196\)
119297
Statements I The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty, is \({ }^9 \mathrm{C}_3\). Statements II The number of ways of choosing any 3 place from 9 different places is \({ }^9 \mathrm{C}_3\).
1 Statement I is true, Statement II is true; Statements II is not a correct explanation of Statement I
2 Statement I is true, Statement II is false
3 Statement I is false, Statement II is true
4 Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
Explanation:
A Statement (I) Given, \(\mathrm{n}=10, \mathrm{r}=4\) Required no of ways \(={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{10-1} \mathrm{C}_{4-1}\) \(={ }^9 \mathrm{C}_3\) Statement (II) there are 9 different places. Required no of ways \(={ }^9 \mathrm{C}_3\) Hence both statement are correct but \(\mathrm{II}^{\text {nd }}\) statement is not a correct explanation of \(\mathrm{I}^{\text {st }}\) statement.
AIEEE-2011
Permutation and Combination
119298
There are 10 points in a plane, out of these 6 are collinear. If \(\mathbf{N}\) is the number of triangles formed by joining these points, then
1 \(\mathrm{N}>190\)
2 \(\mathrm{N} \leq 100\)
3 \(100\lt \mathrm{N} \leq 140\)
4 \(140\lt \mathrm{N} \leq 190\)
Explanation:
B Given, \(\mathrm{n}=10\) points, \(\mathrm{m}=6\) If out of \(n\) points \(m\) are collinear then no of triangles \(={ }^{\mathrm{n}} \mathrm{C}_3-{ }^{\mathrm{m}} \mathrm{C}_3\) \(={ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=\frac{10 !}{3 !(10-3) !}-\frac{6 !}{3 !(6-3) !}\) \(=\frac{10.8 .9 .7 !}{3.2 .17 !}-\frac{6.5 .4 .3 \text { ! }}{3.23 !}\) \(=120-20\) \(=100\)
AIEEE-2011
Permutation and Combination
119300
For which value of \(n \in N, n, i\) has 13 trailing zeros?
1 51
2 54
3 57
4 60
Explanation:
C \(n\) ! exponent of 5 is always less than of equal to exponent of 2 \(13=\frac{\mathrm{n}}{5}+\frac{\mathrm{n}}{5^2}+\frac{\mathrm{n}}{5^3}+\ldots \ldots \ldots .\) According to option \(\mathrm{n}=57\) \(\frac{57}{5}+\frac{57}{25}+\frac{57}{125}\) \(=11+2+\ldots \ldots 0\)Hence in 57 ! the power of 5 is 13
AP EAMCET-23.09.2020
Permutation and Combination
119301
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choice available to him is
1 196
2 140
3 168
4 176
Explanation:
A If we choose 4 then no of choice to choose 4 from 5 is \({ }^5 \mathrm{C}_4\) \({ }^5 \mathrm{C}_4 \times{ }^8 \mathrm{C}_6\) \(\frac{5 !}{4 !(5-4) !} \times \frac{8 !}{6 !(8-6) !}\) \(\frac{5 !}{4 ! 1 !} \times \frac{8.76 . !}{6.1 .2 !}\) \(5 \times 28=140\) Selecting five out of first five question and five out of remaining eight questions \({ }^5 \mathrm{C}_5 \times{ }^8 \mathrm{C}_5\) \(=1 \times \frac{8 !}{5 !(8-5) !}\) \(=1 \times \frac{8 \times 7 \times 6 \times 5 !}{5 ! 3 !}\) \(=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}\) \(=56\)Hence, the total no of choice \(140+96=196\)
119297
Statements I The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty, is \({ }^9 \mathrm{C}_3\). Statements II The number of ways of choosing any 3 place from 9 different places is \({ }^9 \mathrm{C}_3\).
1 Statement I is true, Statement II is true; Statements II is not a correct explanation of Statement I
2 Statement I is true, Statement II is false
3 Statement I is false, Statement II is true
4 Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
Explanation:
A Statement (I) Given, \(\mathrm{n}=10, \mathrm{r}=4\) Required no of ways \(={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{10-1} \mathrm{C}_{4-1}\) \(={ }^9 \mathrm{C}_3\) Statement (II) there are 9 different places. Required no of ways \(={ }^9 \mathrm{C}_3\) Hence both statement are correct but \(\mathrm{II}^{\text {nd }}\) statement is not a correct explanation of \(\mathrm{I}^{\text {st }}\) statement.
AIEEE-2011
Permutation and Combination
119298
There are 10 points in a plane, out of these 6 are collinear. If \(\mathbf{N}\) is the number of triangles formed by joining these points, then
1 \(\mathrm{N}>190\)
2 \(\mathrm{N} \leq 100\)
3 \(100\lt \mathrm{N} \leq 140\)
4 \(140\lt \mathrm{N} \leq 190\)
Explanation:
B Given, \(\mathrm{n}=10\) points, \(\mathrm{m}=6\) If out of \(n\) points \(m\) are collinear then no of triangles \(={ }^{\mathrm{n}} \mathrm{C}_3-{ }^{\mathrm{m}} \mathrm{C}_3\) \(={ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=\frac{10 !}{3 !(10-3) !}-\frac{6 !}{3 !(6-3) !}\) \(=\frac{10.8 .9 .7 !}{3.2 .17 !}-\frac{6.5 .4 .3 \text { ! }}{3.23 !}\) \(=120-20\) \(=100\)
AIEEE-2011
Permutation and Combination
119300
For which value of \(n \in N, n, i\) has 13 trailing zeros?
1 51
2 54
3 57
4 60
Explanation:
C \(n\) ! exponent of 5 is always less than of equal to exponent of 2 \(13=\frac{\mathrm{n}}{5}+\frac{\mathrm{n}}{5^2}+\frac{\mathrm{n}}{5^3}+\ldots \ldots \ldots .\) According to option \(\mathrm{n}=57\) \(\frac{57}{5}+\frac{57}{25}+\frac{57}{125}\) \(=11+2+\ldots \ldots 0\)Hence in 57 ! the power of 5 is 13
AP EAMCET-23.09.2020
Permutation and Combination
119301
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choice available to him is
1 196
2 140
3 168
4 176
Explanation:
A If we choose 4 then no of choice to choose 4 from 5 is \({ }^5 \mathrm{C}_4\) \({ }^5 \mathrm{C}_4 \times{ }^8 \mathrm{C}_6\) \(\frac{5 !}{4 !(5-4) !} \times \frac{8 !}{6 !(8-6) !}\) \(\frac{5 !}{4 ! 1 !} \times \frac{8.76 . !}{6.1 .2 !}\) \(5 \times 28=140\) Selecting five out of first five question and five out of remaining eight questions \({ }^5 \mathrm{C}_5 \times{ }^8 \mathrm{C}_5\) \(=1 \times \frac{8 !}{5 !(8-5) !}\) \(=1 \times \frac{8 \times 7 \times 6 \times 5 !}{5 ! 3 !}\) \(=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}\) \(=56\)Hence, the total no of choice \(140+96=196\)
