NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
118981
The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is
1 271
2 272
3 300
4 None of these
Explanation:
C Any number between 1 to 999 is of form the abc where, \(0 \leq \mathrm{a}, \mathrm{b}, \mathrm{c} \leq 9\). Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in \(1 \times{ }^3 \mathrm{C}_1\) \(\times 9 \times 9=243\) ways, next 5 can occur in exactly two places in \({ }^3 \mathrm{C}_2 \times 9=27\) Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs \(=1 \times 243+27 \times 2+1 \times 3\) \(=243+54+3\) \(=300\)
BITSAT-2017
Permutation and Combination
118982
If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is
1 15
2 25
3 35
4 40
Explanation:
C Given that, Number of identical pencil \(=16\) Number of children \(=4\) If each children gets at least 3 pencils then \(3 \times 4=12\) pencils Left pencil \(=16-12=4\) \(n\) number of pencils can be distributed among \(r\) persons \(={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{4+4-1} \mathrm{C}_{4-1}={ }^7 \mathrm{C}_3\) \(=\frac{7 !}{3 ! 4 !}=\frac{7 \times 6 \times 5}{6}=35\)
BITSAT-2020
Permutation and Combination
118983
The number of possible outcomes in a throw of \(n\) ordinary dice in which at least one of the dice shows an odd number is
1 \(6^{\mathrm{n}}-1\)
2 \(3^{\text {n }}-1\)
3 \(6^{\mathrm{n}}-3^{\mathrm{n}}\)
4 None of these
Explanation:
C We know that, there are total 6 face in a dice so the total number of ways \(=6 \times 6 \times \ldots . n=6^{\text {n }}\) Showing face in every time \(=3\) So, the total number of ways to show only even number \(=3 \times 3 \times \ldots \ldots n=3^{\mathrm{n}}\) \(\therefore\) Required number of ways \(=\) Total face-showing face \(=6^{\mathrm{n}}-3^{\mathrm{n}}\)
BITSAT - 2020
Permutation and Combination
118984
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is
1 \(\frac{(10) !}{6 !}\)
2 \(\frac{(10) !}{24}\)
3 \(\frac{(9) !}{24}\)
4 None of these
Explanation:
B Ways to select 6 people out of 10 people\) \({ }^{10} \mathrm{C}_6\) Permutation of 6 on round table \(=5 !\) Ways to arrange four people at the first \(=(4-1) !\) | Permutation of 4 on round table \(=3\) ! Then, total number of arrangements \(={ }^{10} \mathrm{C}_6 .5 ! .3\) ! \(=\frac{(10) !}{6 ! .4 !} \cdot 5 ! 3 !=\frac{(10) !}{24} \text {. }\)
118981
The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is
1 271
2 272
3 300
4 None of these
Explanation:
C Any number between 1 to 999 is of form the abc where, \(0 \leq \mathrm{a}, \mathrm{b}, \mathrm{c} \leq 9\). Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in \(1 \times{ }^3 \mathrm{C}_1\) \(\times 9 \times 9=243\) ways, next 5 can occur in exactly two places in \({ }^3 \mathrm{C}_2 \times 9=27\) Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs \(=1 \times 243+27 \times 2+1 \times 3\) \(=243+54+3\) \(=300\)
BITSAT-2017
Permutation and Combination
118982
If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is
1 15
2 25
3 35
4 40
Explanation:
C Given that, Number of identical pencil \(=16\) Number of children \(=4\) If each children gets at least 3 pencils then \(3 \times 4=12\) pencils Left pencil \(=16-12=4\) \(n\) number of pencils can be distributed among \(r\) persons \(={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{4+4-1} \mathrm{C}_{4-1}={ }^7 \mathrm{C}_3\) \(=\frac{7 !}{3 ! 4 !}=\frac{7 \times 6 \times 5}{6}=35\)
BITSAT-2020
Permutation and Combination
118983
The number of possible outcomes in a throw of \(n\) ordinary dice in which at least one of the dice shows an odd number is
1 \(6^{\mathrm{n}}-1\)
2 \(3^{\text {n }}-1\)
3 \(6^{\mathrm{n}}-3^{\mathrm{n}}\)
4 None of these
Explanation:
C We know that, there are total 6 face in a dice so the total number of ways \(=6 \times 6 \times \ldots . n=6^{\text {n }}\) Showing face in every time \(=3\) So, the total number of ways to show only even number \(=3 \times 3 \times \ldots \ldots n=3^{\mathrm{n}}\) \(\therefore\) Required number of ways \(=\) Total face-showing face \(=6^{\mathrm{n}}-3^{\mathrm{n}}\)
BITSAT - 2020
Permutation and Combination
118984
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is
1 \(\frac{(10) !}{6 !}\)
2 \(\frac{(10) !}{24}\)
3 \(\frac{(9) !}{24}\)
4 None of these
Explanation:
B Ways to select 6 people out of 10 people\) \({ }^{10} \mathrm{C}_6\) Permutation of 6 on round table \(=5 !\) Ways to arrange four people at the first \(=(4-1) !\) | Permutation of 4 on round table \(=3\) ! Then, total number of arrangements \(={ }^{10} \mathrm{C}_6 .5 ! .3\) ! \(=\frac{(10) !}{6 ! .4 !} \cdot 5 ! 3 !=\frac{(10) !}{24} \text {. }\)
118981
The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is
1 271
2 272
3 300
4 None of these
Explanation:
C Any number between 1 to 999 is of form the abc where, \(0 \leq \mathrm{a}, \mathrm{b}, \mathrm{c} \leq 9\). Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in \(1 \times{ }^3 \mathrm{C}_1\) \(\times 9 \times 9=243\) ways, next 5 can occur in exactly two places in \({ }^3 \mathrm{C}_2 \times 9=27\) Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs \(=1 \times 243+27 \times 2+1 \times 3\) \(=243+54+3\) \(=300\)
BITSAT-2017
Permutation and Combination
118982
If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is
1 15
2 25
3 35
4 40
Explanation:
C Given that, Number of identical pencil \(=16\) Number of children \(=4\) If each children gets at least 3 pencils then \(3 \times 4=12\) pencils Left pencil \(=16-12=4\) \(n\) number of pencils can be distributed among \(r\) persons \(={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{4+4-1} \mathrm{C}_{4-1}={ }^7 \mathrm{C}_3\) \(=\frac{7 !}{3 ! 4 !}=\frac{7 \times 6 \times 5}{6}=35\)
BITSAT-2020
Permutation and Combination
118983
The number of possible outcomes in a throw of \(n\) ordinary dice in which at least one of the dice shows an odd number is
1 \(6^{\mathrm{n}}-1\)
2 \(3^{\text {n }}-1\)
3 \(6^{\mathrm{n}}-3^{\mathrm{n}}\)
4 None of these
Explanation:
C We know that, there are total 6 face in a dice so the total number of ways \(=6 \times 6 \times \ldots . n=6^{\text {n }}\) Showing face in every time \(=3\) So, the total number of ways to show only even number \(=3 \times 3 \times \ldots \ldots n=3^{\mathrm{n}}\) \(\therefore\) Required number of ways \(=\) Total face-showing face \(=6^{\mathrm{n}}-3^{\mathrm{n}}\)
BITSAT - 2020
Permutation and Combination
118984
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is
1 \(\frac{(10) !}{6 !}\)
2 \(\frac{(10) !}{24}\)
3 \(\frac{(9) !}{24}\)
4 None of these
Explanation:
B Ways to select 6 people out of 10 people\) \({ }^{10} \mathrm{C}_6\) Permutation of 6 on round table \(=5 !\) Ways to arrange four people at the first \(=(4-1) !\) | Permutation of 4 on round table \(=3\) ! Then, total number of arrangements \(={ }^{10} \mathrm{C}_6 .5 ! .3\) ! \(=\frac{(10) !}{6 ! .4 !} \cdot 5 ! 3 !=\frac{(10) !}{24} \text {. }\)
118981
The number of times the digit 5 will be written when listing the integers from 1 to 1000 , is
1 271
2 272
3 300
4 None of these
Explanation:
C Any number between 1 to 999 is of form the abc where, \(0 \leq \mathrm{a}, \mathrm{b}, \mathrm{c} \leq 9\). Let us first count the number in which 5 occurs exactly once. Since, 5 can occur at one place in \(1 \times{ }^3 \mathrm{C}_1\) \(\times 9 \times 9=243\) ways, next 5 can occur in exactly two places in \({ }^3 \mathrm{C}_2 \times 9=27\) Lastly, 5 can occur in all three digits in only one way. Hence, the number of times 5 occurs \(=1 \times 243+27 \times 2+1 \times 3\) \(=243+54+3\) \(=300\)
BITSAT-2017
Permutation and Combination
118982
If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is
1 15
2 25
3 35
4 40
Explanation:
C Given that, Number of identical pencil \(=16\) Number of children \(=4\) If each children gets at least 3 pencils then \(3 \times 4=12\) pencils Left pencil \(=16-12=4\) \(n\) number of pencils can be distributed among \(r\) persons \(={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{r}-1}\) \(={ }^{4+4-1} \mathrm{C}_{4-1}={ }^7 \mathrm{C}_3\) \(=\frac{7 !}{3 ! 4 !}=\frac{7 \times 6 \times 5}{6}=35\)
BITSAT-2020
Permutation and Combination
118983
The number of possible outcomes in a throw of \(n\) ordinary dice in which at least one of the dice shows an odd number is
1 \(6^{\mathrm{n}}-1\)
2 \(3^{\text {n }}-1\)
3 \(6^{\mathrm{n}}-3^{\mathrm{n}}\)
4 None of these
Explanation:
C We know that, there are total 6 face in a dice so the total number of ways \(=6 \times 6 \times \ldots . n=6^{\text {n }}\) Showing face in every time \(=3\) So, the total number of ways to show only even number \(=3 \times 3 \times \ldots \ldots n=3^{\mathrm{n}}\) \(\therefore\) Required number of ways \(=\) Total face-showing face \(=6^{\mathrm{n}}-3^{\mathrm{n}}\)
BITSAT - 2020
Permutation and Combination
118984
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is
1 \(\frac{(10) !}{6 !}\)
2 \(\frac{(10) !}{24}\)
3 \(\frac{(9) !}{24}\)
4 None of these
Explanation:
B Ways to select 6 people out of 10 people\) \({ }^{10} \mathrm{C}_6\) Permutation of 6 on round table \(=5 !\) Ways to arrange four people at the first \(=(4-1) !\) | Permutation of 4 on round table \(=3\) ! Then, total number of arrangements \(={ }^{10} \mathrm{C}_6 .5 ! .3\) ! \(=\frac{(10) !}{6 ! .4 !} \cdot 5 ! 3 !=\frac{(10) !}{24} \text {. }\)
