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:
. . . . . . . . . . . . Adding vertically, ..
. . . . . . . . . . . . Adding vertically, ..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:
The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 = 6, A 2 2 = -2, A 2 3 = -14 A 3 1 = 1, A 3 2 = 5, A 3 3 = 3 ..
The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 = 6, A 2 2 = -2, A 2 3 = -14 A 3 1 = 1, A 3 2 = 5, A 3 3 = 3 ..Suggested answer:
We have 3A - 2B = 3A+(-2)..
We have 3A - 2B = 3A+(-2)..Suggested answer:
Let S n = 1+2+3+4+...+n This series is an A.P. Here a=1, d=1, l = t n = n 2. To find the sum to squares of first n natural numbers. or..
Let S n = 1+2+3+4+...+n This series is an A.P. Here a=1, d=1, l = t n = n 2. To find the sum to squares of first n natural numbers. or..Suggested answer:
Subtracting (ii) from (i), we get ..
Subtracting (ii) from (i), we get ..Suggested answer:
Subtracting (ii) from (i), we get ..
Subtracting (ii) from (i), we get ..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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