119195
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together?
1 4 !
2 5 !
3 \(4 !+5\) !
4 4 ! \(\times 5\) !
Explanation:
D Here leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls can sit in 5 ! ways. \(=(5-1) ! \times 5 !\)Hence the required number \(=4 ! \times 5\) !
BITSAT-2013
Permutation and Combination
119198
The total number of numbers not more than 20 digits that are formed by using the digits \(0,1,2\), 3 , and 4 is
1 \(5^{20}\)
2 \(5^{20}-1\)
3 \(5^{20}+1\)
4 None of these
Explanation:
A Here, \(\text { Number of number }=\sum_{\mathrm{r}=1}^{20} \text { (number of } \mathrm{r} \text { digit numbers) }\) \(=5+4.5+4.5^{\mathrm{r}=1}+4.5^3+\ldots . .+4.5^{19}\) \(=5+4 \times 5\left[\frac{5^{19}-1}{5-1}\right]=5+5^{20}-5=5^{20}\)
UPSEE-2010
Permutation and Combination
119203
\(S=\{1,2,3, \ldots .20\}\) is to be partitioned into four sets \(A, B, C\) and \(D\) of equal size. The number of ways it can be done, is
1 \(\frac{20 !}{4 ! \times 5 !}\)
2 \(\frac{20 !}{4^5}\)
3 \(\frac{20 !}{(5 !)^4}\)
4 \(\frac{20 !}{(4 !)^5}\)
Explanation:
C Set \(S=\{1,2,3, \ldots, 20\}\) is to be partition into four set of equal size i.e. having 5-5 numbers. The number of ways in which it can be done \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times{ }^5 \mathrm{C}_5\) \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times 1\) \(=\frac{20 !}{15 ! \times 5 !} \times \frac{15 !}{10 ! \times 5 !} \times \frac{10 !}{5 ! \times 5 !} \times 1=\frac{(20 !)}{(5 !)^4}\)
Manipal UGET-2014
Permutation and Combination
119211
The total number of way in which 30 books can be distributed among 5 students is
1 \({ }^{30} \mathrm{C}_5\)
2 \({ }^{34} \mathrm{C}_5\)
3 \({ }^{30} \mathrm{C}_4\)
4 \({ }^{34} \mathrm{C}_4\)
Explanation:
A Given that, \(\text {Number of books} =30\) \(\text {Number of student} =5\) \(So, \text {the number of ways} ={ }^{30} \mathrm{C}_5\)
119195
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together?
1 4 !
2 5 !
3 \(4 !+5\) !
4 4 ! \(\times 5\) !
Explanation:
D Here leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls can sit in 5 ! ways. \(=(5-1) ! \times 5 !\)Hence the required number \(=4 ! \times 5\) !
BITSAT-2013
Permutation and Combination
119198
The total number of numbers not more than 20 digits that are formed by using the digits \(0,1,2\), 3 , and 4 is
1 \(5^{20}\)
2 \(5^{20}-1\)
3 \(5^{20}+1\)
4 None of these
Explanation:
A Here, \(\text { Number of number }=\sum_{\mathrm{r}=1}^{20} \text { (number of } \mathrm{r} \text { digit numbers) }\) \(=5+4.5+4.5^{\mathrm{r}=1}+4.5^3+\ldots . .+4.5^{19}\) \(=5+4 \times 5\left[\frac{5^{19}-1}{5-1}\right]=5+5^{20}-5=5^{20}\)
UPSEE-2010
Permutation and Combination
119203
\(S=\{1,2,3, \ldots .20\}\) is to be partitioned into four sets \(A, B, C\) and \(D\) of equal size. The number of ways it can be done, is
1 \(\frac{20 !}{4 ! \times 5 !}\)
2 \(\frac{20 !}{4^5}\)
3 \(\frac{20 !}{(5 !)^4}\)
4 \(\frac{20 !}{(4 !)^5}\)
Explanation:
C Set \(S=\{1,2,3, \ldots, 20\}\) is to be partition into four set of equal size i.e. having 5-5 numbers. The number of ways in which it can be done \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times{ }^5 \mathrm{C}_5\) \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times 1\) \(=\frac{20 !}{15 ! \times 5 !} \times \frac{15 !}{10 ! \times 5 !} \times \frac{10 !}{5 ! \times 5 !} \times 1=\frac{(20 !)}{(5 !)^4}\)
Manipal UGET-2014
Permutation and Combination
119211
The total number of way in which 30 books can be distributed among 5 students is
1 \({ }^{30} \mathrm{C}_5\)
2 \({ }^{34} \mathrm{C}_5\)
3 \({ }^{30} \mathrm{C}_4\)
4 \({ }^{34} \mathrm{C}_4\)
Explanation:
A Given that, \(\text {Number of books} =30\) \(\text {Number of student} =5\) \(So, \text {the number of ways} ={ }^{30} \mathrm{C}_5\)
119195
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together?
1 4 !
2 5 !
3 \(4 !+5\) !
4 4 ! \(\times 5\) !
Explanation:
D Here leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls can sit in 5 ! ways. \(=(5-1) ! \times 5 !\)Hence the required number \(=4 ! \times 5\) !
BITSAT-2013
Permutation and Combination
119198
The total number of numbers not more than 20 digits that are formed by using the digits \(0,1,2\), 3 , and 4 is
1 \(5^{20}\)
2 \(5^{20}-1\)
3 \(5^{20}+1\)
4 None of these
Explanation:
A Here, \(\text { Number of number }=\sum_{\mathrm{r}=1}^{20} \text { (number of } \mathrm{r} \text { digit numbers) }\) \(=5+4.5+4.5^{\mathrm{r}=1}+4.5^3+\ldots . .+4.5^{19}\) \(=5+4 \times 5\left[\frac{5^{19}-1}{5-1}\right]=5+5^{20}-5=5^{20}\)
UPSEE-2010
Permutation and Combination
119203
\(S=\{1,2,3, \ldots .20\}\) is to be partitioned into four sets \(A, B, C\) and \(D\) of equal size. The number of ways it can be done, is
1 \(\frac{20 !}{4 ! \times 5 !}\)
2 \(\frac{20 !}{4^5}\)
3 \(\frac{20 !}{(5 !)^4}\)
4 \(\frac{20 !}{(4 !)^5}\)
Explanation:
C Set \(S=\{1,2,3, \ldots, 20\}\) is to be partition into four set of equal size i.e. having 5-5 numbers. The number of ways in which it can be done \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times{ }^5 \mathrm{C}_5\) \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times 1\) \(=\frac{20 !}{15 ! \times 5 !} \times \frac{15 !}{10 ! \times 5 !} \times \frac{10 !}{5 ! \times 5 !} \times 1=\frac{(20 !)}{(5 !)^4}\)
Manipal UGET-2014
Permutation and Combination
119211
The total number of way in which 30 books can be distributed among 5 students is
1 \({ }^{30} \mathrm{C}_5\)
2 \({ }^{34} \mathrm{C}_5\)
3 \({ }^{30} \mathrm{C}_4\)
4 \({ }^{34} \mathrm{C}_4\)
Explanation:
A Given that, \(\text {Number of books} =30\) \(\text {Number of student} =5\) \(So, \text {the number of ways} ={ }^{30} \mathrm{C}_5\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
119195
In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together?
1 4 !
2 5 !
3 \(4 !+5\) !
4 4 ! \(\times 5\) !
Explanation:
D Here leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls can sit in 5 ! ways. \(=(5-1) ! \times 5 !\)Hence the required number \(=4 ! \times 5\) !
BITSAT-2013
Permutation and Combination
119198
The total number of numbers not more than 20 digits that are formed by using the digits \(0,1,2\), 3 , and 4 is
1 \(5^{20}\)
2 \(5^{20}-1\)
3 \(5^{20}+1\)
4 None of these
Explanation:
A Here, \(\text { Number of number }=\sum_{\mathrm{r}=1}^{20} \text { (number of } \mathrm{r} \text { digit numbers) }\) \(=5+4.5+4.5^{\mathrm{r}=1}+4.5^3+\ldots . .+4.5^{19}\) \(=5+4 \times 5\left[\frac{5^{19}-1}{5-1}\right]=5+5^{20}-5=5^{20}\)
UPSEE-2010
Permutation and Combination
119203
\(S=\{1,2,3, \ldots .20\}\) is to be partitioned into four sets \(A, B, C\) and \(D\) of equal size. The number of ways it can be done, is
1 \(\frac{20 !}{4 ! \times 5 !}\)
2 \(\frac{20 !}{4^5}\)
3 \(\frac{20 !}{(5 !)^4}\)
4 \(\frac{20 !}{(4 !)^5}\)
Explanation:
C Set \(S=\{1,2,3, \ldots, 20\}\) is to be partition into four set of equal size i.e. having 5-5 numbers. The number of ways in which it can be done \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times{ }^5 \mathrm{C}_5\) \(={ }^{20} \mathrm{C}_5 \times{ }^{15} \mathrm{C}_5 \times{ }^{10} \mathrm{C}_5 \times 1\) \(=\frac{20 !}{15 ! \times 5 !} \times \frac{15 !}{10 ! \times 5 !} \times \frac{10 !}{5 ! \times 5 !} \times 1=\frac{(20 !)}{(5 !)^4}\)
Manipal UGET-2014
Permutation and Combination
119211
The total number of way in which 30 books can be distributed among 5 students is
1 \({ }^{30} \mathrm{C}_5\)
2 \({ }^{34} \mathrm{C}_5\)
3 \({ }^{30} \mathrm{C}_4\)
4 \({ }^{34} \mathrm{C}_4\)
Explanation:
A Given that, \(\text {Number of books} =30\) \(\text {Number of student} =5\) \(So, \text {the number of ways} ={ }^{30} \mathrm{C}_5\)
