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Discrete Mathematics - Test Questions I
Question 1 - Question: From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done? Answer: We are to select 4 students from 32. This selection can done ..
Question 1 - Question: From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done? Answer: We are to select 4 students from 32. This selection can done ..Suggested answer:
From (1) and (2) (A + B) + C = A + (B + C) verify the associative la..
From (1) and (2) (A + B) + C = A + (B + C) verify the associative la..Suggested answer:
4. Method of finding sum of a series whose n t h term is know..
4. Method of finding sum of a series whose n t h term is know..Suggested answer:
n t h term of (1,2,3,....) is n Subtracting (ii) from (i), we get ..
n t h term of (1,2,3,....) is n Subtracting (ii) from (i), we get ..Suggested answer:
i) 1.2 + 2.4 + 3.8 +.... to n terms The n t h term of (1,2,3,....n) is n. The n t h term of (2,4,8,...) is The n t h term of the given series is n2 n Subtracting (ii) from (i), we have ..
i) 1.2 + 2.4 + 3.8 +.... to n terms The n t h term of (1,2,3,....n) is n. The n t h term of (2,4,8,...) is The n t h term of the given series is n2 n Subtracting (ii) from (i), we have ..Suggested answer:
From (1) and (2), (A + B) + C = A + (B + C) \ The associative law is verifi..
From (1) and (2), (A + B) + C = A + (B + C) \ The associative law is verifi..Suggested answer:
We are to select 4 students from 32. This selection can done i..
We are to select 4 students from 32. This selection can done i..Suggested answer:
Let the n t h term of series be an 3 + bn 2 + cn + d t n = an 3 + bn 2 + cn + d where a, b, c, d are constants. ..
Let the n t h term of series be an 3 + bn 2 + cn + d t n = an 3 + bn 2 + cn + d where a, b, c, d are constants. ..Suggested answer:
The arrangement is as shown in the figure, the boy X will have B 2 , B 3 as neighbours. The girl Y will have G 2 , G 3 as neighbours. The two boys B 2 , B 3 can be arranged in two ways. The two girls G 2 , G 3 can be arranged in two ways. Hence, the total number of arrangements = 2 x 2 = ..
The arrangement is as shown in the figure, the boy X will have B 2 , B 3 as neighbours. The girl Y will have G 2 , G 3 as neighbours. The two boys B 2 , B 3 can be arranged in two ways. The two girls G 2 , G 3 can be arranged in two ways. Hence, the total number of arrangements = 2 x 2 = .. Result
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