119192
A bag contains 3 one rupee coins, 4 fifty paisa coins and 5 ten paisa coins. How many selection of money can be formed by taking atleast one coin from the bag.
1 90
2 140
3 119
4 125
Explanation:
C Given that Number of coins of one rupee \(=3\) Number of coins of 50 paisa \(=4\) Number of coins of 10 paisa \(=5\) \(=(3+1)(4+1)(5+1)-1=120-1=119\)
BITSAT-2005
Permutation and Combination
119193
With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is
1 5780
2 2890
3 5440
4 2720
Explanation:
A Given that, Number of constant \(=17\) Number of vowels \(=5\) Number of way of selecting vowels \(={ }^5 \mathrm{C}_2\) \(\frac{5 !}{3 ! 2 !}=10\) Number of way of selecting consonants \(={ }^{17} \mathrm{C}_1 \times{ }^{17} \mathrm{C}_1\) \(\frac{17 !}{16 !} \times \frac{17 !}{16 !}=289\) We can use repeated consonant at the end. Total number of ways to select consonant \(=2 \times 289=578\) Hence, total number of ways \(=10 \times 578 \quad \text { (According to product rule) }\) \(=5780\)(According to product rule)
BITSAT-2010
Permutation and Combination
119194
In how many ways can a committee of 5 be made out 6 men and 4 women containing atleast one woman?
1 246
2 222
3 186
4 None of these
Explanation:
A Given that, The number of men \(=6\) The number of women \(=4\) Committee can be formed in the following ways Case I :- 4 men and 1 women \(={ }^4 \mathrm{C}_1 \times{ }^6 \mathrm{C}_4\) \(=\frac{4 !}{1 ! \times 3 !} \times \frac{6 !}{4 ! 2 !}=60\) Case II :- 3 men and 2 woman \(={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2\) \(=\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}=120\) Case III :- 2 men and 3 woman \(={ }^6 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3=\frac{4 !}{3 ! 1 !} \times \frac{6 !}{2 ! 4 !}\) \(=60\) Case IV :- 1 men and 4 woman \(={ }^6 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4\) \(=\frac{6 !}{1 ! 5 !} \times \frac{4 !}{4 ! 0 !}=6\) So, the require number of ways \(=\) sum of all case \(=60+120+60+6=246\)
BITSAT-2014
Permutation and Combination
119196
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if atleast one black ball is to be included in the draw?
1 64
2 129
3 84
4 None of these
Explanation:
A At least one black ball can be drawn in the following ways Case I: One black and two other colour balls \(={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2=3 \times 15=45\) Case II: Two black and one other colour balls \(={ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1=3 \times 6=18\) Case III: All the three are black \(={ }^3 \mathrm{C}_3 \times{ }^6 \mathrm{C}_0=1\) \(\therefore\) Required no. of ways \(=45+18+1=64\)
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Permutation and Combination
119192
A bag contains 3 one rupee coins, 4 fifty paisa coins and 5 ten paisa coins. How many selection of money can be formed by taking atleast one coin from the bag.
1 90
2 140
3 119
4 125
Explanation:
C Given that Number of coins of one rupee \(=3\) Number of coins of 50 paisa \(=4\) Number of coins of 10 paisa \(=5\) \(=(3+1)(4+1)(5+1)-1=120-1=119\)
BITSAT-2005
Permutation and Combination
119193
With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is
1 5780
2 2890
3 5440
4 2720
Explanation:
A Given that, Number of constant \(=17\) Number of vowels \(=5\) Number of way of selecting vowels \(={ }^5 \mathrm{C}_2\) \(\frac{5 !}{3 ! 2 !}=10\) Number of way of selecting consonants \(={ }^{17} \mathrm{C}_1 \times{ }^{17} \mathrm{C}_1\) \(\frac{17 !}{16 !} \times \frac{17 !}{16 !}=289\) We can use repeated consonant at the end. Total number of ways to select consonant \(=2 \times 289=578\) Hence, total number of ways \(=10 \times 578 \quad \text { (According to product rule) }\) \(=5780\)(According to product rule)
BITSAT-2010
Permutation and Combination
119194
In how many ways can a committee of 5 be made out 6 men and 4 women containing atleast one woman?
1 246
2 222
3 186
4 None of these
Explanation:
A Given that, The number of men \(=6\) The number of women \(=4\) Committee can be formed in the following ways Case I :- 4 men and 1 women \(={ }^4 \mathrm{C}_1 \times{ }^6 \mathrm{C}_4\) \(=\frac{4 !}{1 ! \times 3 !} \times \frac{6 !}{4 ! 2 !}=60\) Case II :- 3 men and 2 woman \(={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2\) \(=\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}=120\) Case III :- 2 men and 3 woman \(={ }^6 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3=\frac{4 !}{3 ! 1 !} \times \frac{6 !}{2 ! 4 !}\) \(=60\) Case IV :- 1 men and 4 woman \(={ }^6 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4\) \(=\frac{6 !}{1 ! 5 !} \times \frac{4 !}{4 ! 0 !}=6\) So, the require number of ways \(=\) sum of all case \(=60+120+60+6=246\)
BITSAT-2014
Permutation and Combination
119196
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if atleast one black ball is to be included in the draw?
1 64
2 129
3 84
4 None of these
Explanation:
A At least one black ball can be drawn in the following ways Case I: One black and two other colour balls \(={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2=3 \times 15=45\) Case II: Two black and one other colour balls \(={ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1=3 \times 6=18\) Case III: All the three are black \(={ }^3 \mathrm{C}_3 \times{ }^6 \mathrm{C}_0=1\) \(\therefore\) Required no. of ways \(=45+18+1=64\)
119192
A bag contains 3 one rupee coins, 4 fifty paisa coins and 5 ten paisa coins. How many selection of money can be formed by taking atleast one coin from the bag.
1 90
2 140
3 119
4 125
Explanation:
C Given that Number of coins of one rupee \(=3\) Number of coins of 50 paisa \(=4\) Number of coins of 10 paisa \(=5\) \(=(3+1)(4+1)(5+1)-1=120-1=119\)
BITSAT-2005
Permutation and Combination
119193
With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is
1 5780
2 2890
3 5440
4 2720
Explanation:
A Given that, Number of constant \(=17\) Number of vowels \(=5\) Number of way of selecting vowels \(={ }^5 \mathrm{C}_2\) \(\frac{5 !}{3 ! 2 !}=10\) Number of way of selecting consonants \(={ }^{17} \mathrm{C}_1 \times{ }^{17} \mathrm{C}_1\) \(\frac{17 !}{16 !} \times \frac{17 !}{16 !}=289\) We can use repeated consonant at the end. Total number of ways to select consonant \(=2 \times 289=578\) Hence, total number of ways \(=10 \times 578 \quad \text { (According to product rule) }\) \(=5780\)(According to product rule)
BITSAT-2010
Permutation and Combination
119194
In how many ways can a committee of 5 be made out 6 men and 4 women containing atleast one woman?
1 246
2 222
3 186
4 None of these
Explanation:
A Given that, The number of men \(=6\) The number of women \(=4\) Committee can be formed in the following ways Case I :- 4 men and 1 women \(={ }^4 \mathrm{C}_1 \times{ }^6 \mathrm{C}_4\) \(=\frac{4 !}{1 ! \times 3 !} \times \frac{6 !}{4 ! 2 !}=60\) Case II :- 3 men and 2 woman \(={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2\) \(=\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}=120\) Case III :- 2 men and 3 woman \(={ }^6 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3=\frac{4 !}{3 ! 1 !} \times \frac{6 !}{2 ! 4 !}\) \(=60\) Case IV :- 1 men and 4 woman \(={ }^6 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4\) \(=\frac{6 !}{1 ! 5 !} \times \frac{4 !}{4 ! 0 !}=6\) So, the require number of ways \(=\) sum of all case \(=60+120+60+6=246\)
BITSAT-2014
Permutation and Combination
119196
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if atleast one black ball is to be included in the draw?
1 64
2 129
3 84
4 None of these
Explanation:
A At least one black ball can be drawn in the following ways Case I: One black and two other colour balls \(={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2=3 \times 15=45\) Case II: Two black and one other colour balls \(={ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1=3 \times 6=18\) Case III: All the three are black \(={ }^3 \mathrm{C}_3 \times{ }^6 \mathrm{C}_0=1\) \(\therefore\) Required no. of ways \(=45+18+1=64\)
119192
A bag contains 3 one rupee coins, 4 fifty paisa coins and 5 ten paisa coins. How many selection of money can be formed by taking atleast one coin from the bag.
1 90
2 140
3 119
4 125
Explanation:
C Given that Number of coins of one rupee \(=3\) Number of coins of 50 paisa \(=4\) Number of coins of 10 paisa \(=5\) \(=(3+1)(4+1)(5+1)-1=120-1=119\)
BITSAT-2005
Permutation and Combination
119193
With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is
1 5780
2 2890
3 5440
4 2720
Explanation:
A Given that, Number of constant \(=17\) Number of vowels \(=5\) Number of way of selecting vowels \(={ }^5 \mathrm{C}_2\) \(\frac{5 !}{3 ! 2 !}=10\) Number of way of selecting consonants \(={ }^{17} \mathrm{C}_1 \times{ }^{17} \mathrm{C}_1\) \(\frac{17 !}{16 !} \times \frac{17 !}{16 !}=289\) We can use repeated consonant at the end. Total number of ways to select consonant \(=2 \times 289=578\) Hence, total number of ways \(=10 \times 578 \quad \text { (According to product rule) }\) \(=5780\)(According to product rule)
BITSAT-2010
Permutation and Combination
119194
In how many ways can a committee of 5 be made out 6 men and 4 women containing atleast one woman?
1 246
2 222
3 186
4 None of these
Explanation:
A Given that, The number of men \(=6\) The number of women \(=4\) Committee can be formed in the following ways Case I :- 4 men and 1 women \(={ }^4 \mathrm{C}_1 \times{ }^6 \mathrm{C}_4\) \(=\frac{4 !}{1 ! \times 3 !} \times \frac{6 !}{4 ! 2 !}=60\) Case II :- 3 men and 2 woman \(={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2\) \(=\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}=120\) Case III :- 2 men and 3 woman \(={ }^6 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3=\frac{4 !}{3 ! 1 !} \times \frac{6 !}{2 ! 4 !}\) \(=60\) Case IV :- 1 men and 4 woman \(={ }^6 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4\) \(=\frac{6 !}{1 ! 5 !} \times \frac{4 !}{4 ! 0 !}=6\) So, the require number of ways \(=\) sum of all case \(=60+120+60+6=246\)
BITSAT-2014
Permutation and Combination
119196
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if atleast one black ball is to be included in the draw?
1 64
2 129
3 84
4 None of these
Explanation:
A At least one black ball can be drawn in the following ways Case I: One black and two other colour balls \(={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2=3 \times 15=45\) Case II: Two black and one other colour balls \(={ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1=3 \times 6=18\) Case III: All the three are black \(={ }^3 \mathrm{C}_3 \times{ }^6 \mathrm{C}_0=1\) \(\therefore\) Required no. of ways \(=45+18+1=64\)