119003
The number of ways in which 5 boys and 5 girls can be seated for a photograph, so that no two girls sit next to each other is
1 \(6 ! 5\) !
2 \((5 !)^2\)
3 \(\frac{10 !}{(5 !)}\)
4 \(\frac{10 !}{(5 !)^2}\)
Explanation:
A : Given that, number of boys \(=5\) Number of girls \(=5\) The number of ways boys can be arranged = 5! Here 6 places for girls where no two girls are together So, the number of ways, the girls can be arranged = \(6 !\) Hence, the required number of ways \(=5 ! \times 6 !\) \(=6 ! \times 5 !\)
BCECE-2012
Permutation and Combination
119005
Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls.
C : Given that, number of white ball \(=5 \text {, number }\) \(\text { of green balls }=6\) \(\text { Total number of balls }=11\) \(\text { Therefore, total number of cases }={ }^{11} \mathrm{C}_7\) \(\text { The favorable cases },{ }^5 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_4={ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2\) \(\therefore \quad \text { Required probability } =\frac{\text { Favorablecases }}{\text { Totalnumber of cases }}\) \(=\frac{{ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2}{{ }^{11} \mathrm{C}_7}\)
BCECE-2008
Permutation and Combination
119006
The number of natural numbers less than 1000 , in which no two digits are repeated is
1 738
2 792
3 837
4 720
Explanation:
A For number of one digit \(=9\) Number of two digits non repeated number \(={ }^{10} \mathrm{P}_2-{ }^9 \mathrm{P}_1\) \(=90-9=81\) Number of three digit non repeated number \(={ }^{10} \mathrm{P}_3-{ }^9 \mathrm{P}_2\) \(=10 \times 9 \times 8-9 \times 8\) \(=720-72=648\) Hence, the total number of ways \(=9+81+648=738\)
BCECE-2008
Permutation and Combination
119007
For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is
1 \(\frac{2}{7}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{4}{9}\)
Explanation:
B Given that, number of invited guest \(=8\) Total number of way of sitting \(=9\) ! Number of favorable ways of sitting \(=2 \times 8\) ! \(\therefore\) Required probability \(=\frac{\text { Number of event }}{\text { totalevent }}\) \(=\frac{2 \times 8 !}{9 !}=\frac{2}{9}\)
119003
The number of ways in which 5 boys and 5 girls can be seated for a photograph, so that no two girls sit next to each other is
1 \(6 ! 5\) !
2 \((5 !)^2\)
3 \(\frac{10 !}{(5 !)}\)
4 \(\frac{10 !}{(5 !)^2}\)
Explanation:
A : Given that, number of boys \(=5\) Number of girls \(=5\) The number of ways boys can be arranged = 5! Here 6 places for girls where no two girls are together So, the number of ways, the girls can be arranged = \(6 !\) Hence, the required number of ways \(=5 ! \times 6 !\) \(=6 ! \times 5 !\)
BCECE-2012
Permutation and Combination
119005
Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls.
C : Given that, number of white ball \(=5 \text {, number }\) \(\text { of green balls }=6\) \(\text { Total number of balls }=11\) \(\text { Therefore, total number of cases }={ }^{11} \mathrm{C}_7\) \(\text { The favorable cases },{ }^5 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_4={ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2\) \(\therefore \quad \text { Required probability } =\frac{\text { Favorablecases }}{\text { Totalnumber of cases }}\) \(=\frac{{ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2}{{ }^{11} \mathrm{C}_7}\)
BCECE-2008
Permutation and Combination
119006
The number of natural numbers less than 1000 , in which no two digits are repeated is
1 738
2 792
3 837
4 720
Explanation:
A For number of one digit \(=9\) Number of two digits non repeated number \(={ }^{10} \mathrm{P}_2-{ }^9 \mathrm{P}_1\) \(=90-9=81\) Number of three digit non repeated number \(={ }^{10} \mathrm{P}_3-{ }^9 \mathrm{P}_2\) \(=10 \times 9 \times 8-9 \times 8\) \(=720-72=648\) Hence, the total number of ways \(=9+81+648=738\)
BCECE-2008
Permutation and Combination
119007
For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is
1 \(\frac{2}{7}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{4}{9}\)
Explanation:
B Given that, number of invited guest \(=8\) Total number of way of sitting \(=9\) ! Number of favorable ways of sitting \(=2 \times 8\) ! \(\therefore\) Required probability \(=\frac{\text { Number of event }}{\text { totalevent }}\) \(=\frac{2 \times 8 !}{9 !}=\frac{2}{9}\)
119003
The number of ways in which 5 boys and 5 girls can be seated for a photograph, so that no two girls sit next to each other is
1 \(6 ! 5\) !
2 \((5 !)^2\)
3 \(\frac{10 !}{(5 !)}\)
4 \(\frac{10 !}{(5 !)^2}\)
Explanation:
A : Given that, number of boys \(=5\) Number of girls \(=5\) The number of ways boys can be arranged = 5! Here 6 places for girls where no two girls are together So, the number of ways, the girls can be arranged = \(6 !\) Hence, the required number of ways \(=5 ! \times 6 !\) \(=6 ! \times 5 !\)
BCECE-2012
Permutation and Combination
119005
Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls.
C : Given that, number of white ball \(=5 \text {, number }\) \(\text { of green balls }=6\) \(\text { Total number of balls }=11\) \(\text { Therefore, total number of cases }={ }^{11} \mathrm{C}_7\) \(\text { The favorable cases },{ }^5 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_4={ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2\) \(\therefore \quad \text { Required probability } =\frac{\text { Favorablecases }}{\text { Totalnumber of cases }}\) \(=\frac{{ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2}{{ }^{11} \mathrm{C}_7}\)
BCECE-2008
Permutation and Combination
119006
The number of natural numbers less than 1000 , in which no two digits are repeated is
1 738
2 792
3 837
4 720
Explanation:
A For number of one digit \(=9\) Number of two digits non repeated number \(={ }^{10} \mathrm{P}_2-{ }^9 \mathrm{P}_1\) \(=90-9=81\) Number of three digit non repeated number \(={ }^{10} \mathrm{P}_3-{ }^9 \mathrm{P}_2\) \(=10 \times 9 \times 8-9 \times 8\) \(=720-72=648\) Hence, the total number of ways \(=9+81+648=738\)
BCECE-2008
Permutation and Combination
119007
For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is
1 \(\frac{2}{7}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{4}{9}\)
Explanation:
B Given that, number of invited guest \(=8\) Total number of way of sitting \(=9\) ! Number of favorable ways of sitting \(=2 \times 8\) ! \(\therefore\) Required probability \(=\frac{\text { Number of event }}{\text { totalevent }}\) \(=\frac{2 \times 8 !}{9 !}=\frac{2}{9}\)
119003
The number of ways in which 5 boys and 5 girls can be seated for a photograph, so that no two girls sit next to each other is
1 \(6 ! 5\) !
2 \((5 !)^2\)
3 \(\frac{10 !}{(5 !)}\)
4 \(\frac{10 !}{(5 !)^2}\)
Explanation:
A : Given that, number of boys \(=5\) Number of girls \(=5\) The number of ways boys can be arranged = 5! Here 6 places for girls where no two girls are together So, the number of ways, the girls can be arranged = \(6 !\) Hence, the required number of ways \(=5 ! \times 6 !\) \(=6 ! \times 5 !\)
BCECE-2012
Permutation and Combination
119005
Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls.
C : Given that, number of white ball \(=5 \text {, number }\) \(\text { of green balls }=6\) \(\text { Total number of balls }=11\) \(\text { Therefore, total number of cases }={ }^{11} \mathrm{C}_7\) \(\text { The favorable cases },{ }^5 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_4={ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2\) \(\therefore \quad \text { Required probability } =\frac{\text { Favorablecases }}{\text { Totalnumber of cases }}\) \(=\frac{{ }^5 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_2}{{ }^{11} \mathrm{C}_7}\)
BCECE-2008
Permutation and Combination
119006
The number of natural numbers less than 1000 , in which no two digits are repeated is
1 738
2 792
3 837
4 720
Explanation:
A For number of one digit \(=9\) Number of two digits non repeated number \(={ }^{10} \mathrm{P}_2-{ }^9 \mathrm{P}_1\) \(=90-9=81\) Number of three digit non repeated number \(={ }^{10} \mathrm{P}_3-{ }^9 \mathrm{P}_2\) \(=10 \times 9 \times 8-9 \times 8\) \(=720-72=648\) Hence, the total number of ways \(=9+81+648=738\)
BCECE-2008
Permutation and Combination
119007
For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is
1 \(\frac{2}{7}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{4}{9}\)
Explanation:
B Given that, number of invited guest \(=8\) Total number of way of sitting \(=9\) ! Number of favorable ways of sitting \(=2 \times 8\) ! \(\therefore\) Required probability \(=\frac{\text { Number of event }}{\text { totalevent }}\) \(=\frac{2 \times 8 !}{9 !}=\frac{2}{9}\)
