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:
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:
The determinant of coefficients ..
The determinant of coefficients ..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:
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:
Adding vertically, ..
Adding vertically, ..Suggested answer:
Subtracting (ii) from (i), we get ..
Subtracting (ii) from (i), we get ..Suggested answer:
= (14 - 12) - (7 - 3) + (4 - 2) = 2 - 4 + 2 = 0 The system may have infinite number of solutions or no solution. Put x = k in (1) and (2) and solve y + z = 6 - k 2y + 3z = 14 - k. Solving the above two equations, we have z = k + 2 and y = 4 - 2k When x = k, substituting these values of x, y ..
= (14 - 12) - (7 - 3) + (4 - 2) = 2 - 4 + 2 = 0 The system may have infinite number of solutions or no solution. Put x = k in (1) and (2) and solve y + z = 6 - k 2y + 3z = 14 - k. Solving the above two equations, we have z = k + 2 and y = 4 - 2k When x = k, substituting these values of x, y ..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. ..See what our Users say :
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