TCS
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Numerical Ability
Permutation and Combination
A question paper consists of 4 sections with 7 questions in each section. A candidate has to select 2 sections and has to solve 9 questions choosing at least 3 from each of the selected sections. In how many ways can he answer the paper?
Read Solution (Total 10)
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- 4C2*(7C3*7C6+7C4*7C5+7C5*7C4+7C6*7C3)
- 6 years agoHelpfull: Yes(16) No(0)
- At first we need to select 2 sections out of 4 sections=4C2
then we need to answer total 9 questions with atleast 3 questions from each of selected section=(7C3*7C6)+(7C4*7C5)+(7C5*7C4)+(7C6*7C3)
so no of ways we can answer the question paper=(4C2)*((7C3*7C6)+(7C4*7C5)+(7C5*7C4)+(7C6*7C3))=11760. - 6 years agoHelpfull: Yes(9) No(0)
- 4C2*7C3*7C6
- 6 years agoHelpfull: Yes(3) No(16)
- ans is 11760
since in 4 section 2 has to be selected so 4c2 and in 7 atleast 3 has to selected i.e 3 or above 3 so if select 3 in one section second has to be 6 7c3*7c6 and if 4 is selected another should be 5 so 7c4*7c5 and if 5 means another has to 5 so 7c5*7c3 and if 6 another is 3 7c6*7c3 so 4c2*(7c3*7c6+7c4*7c5+7c5*7c4+7c6*7c3)=11760 - 6 years agoHelpfull: Yes(2) No(0)
- 2 sections can be chosen in 4C2 ways and can be solved in 4 ways ((4,5),(5,4),(3,6),(6,3)) total ways = 24
- 6 years agoHelpfull: Yes(1) No(7)
- Given total number of sections = 4.
A candidate has to select 2 sections.
This can happen in C(4,2) Ways.
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Given Total number of Questions in each section = 7.
That means total number of questions in 2 sections = 14.
In the given question he has to solve 9 questions choosing at least 3 from 2 sections.
(1) He can select 3 questions from section 1 and 6 questions from section 2.
= > C(7,3) * C(7,6)
(2) He can select 3 questions from section 2 and 6 questions from section 1.
= > C(7,3) * C(7,6)
(3) He can select 4 questions from section 1 and 5 questions from section 2.
= > C(7,4) * C(7,5)
(4) He can select 5 questions from section 2 and 4 from section 1.
= > C(7,5) * C(7,4).
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Therefore, total number of ways.
= > C(4,2) * (C(7,3) * C(7,6) + C(7,6) * C(7,3) + C(7,5) * C(7,4) + C(7,5) * C(7,4))
= > 6 * (35 * 7 + 7 * 35 + 21 * 35 + 21 * 35)
= > 6 * 1960
= > 11760. - 5 years agoHelpfull: Yes(1) No(0)
- 4c2*7c3*7c6
- 6 years agoHelpfull: Yes(0) No(4)
- =4C2*((7C3*7C6)+(7C4+7C5)+(7C5*7C4)+(7C6+7C3))
- 6 years agoHelpfull: Yes(0) No(0)
- (7c3*7c6)+(7c6*7c3)+(7c4*7c5)+(7c5*7c4)=1540
- 6 years agoHelpfull: Yes(0) No(0)
- 4c2(7c3*7c6+7c4*7c5+7c5*7c4+7c6*7c3)=
- 6 years agoHelpfull: Yes(0) No(0)
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