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# Discrete Mathematics Questions and Answers

Discrete Mathematics Question Bank with Answer:
Discrete maths are maintained as a part of mathematics subject, which is essential for any beginner of studies. Discrete maths cover the topic like number system, element of set theory, algebraic equation relation and functions,Probability, graph theory, and matrix. The question bank is prepared based on the availability  and application of math topic. Required number of questions are given with key answers. By practicing Discrete mathematics question and answers makes the learner easy and at the same time enjoy towards the subject.

1. Find the value of 5.123x10-2 + 6.042x10-2

Solution: 5.123 + 6.042 = 11.165x10-2
the value can be divided by 10 which equals to 1.1165x10-1

2. Insert the missing number 5,20,10,_,20,80

Solution: 5,20,10,_,20,80
Here first digit is multiplied by 4,for the obtained number (second digit) divided by 2,these is how the series follows.
5 x 4 = 20
20 ÷ 2 = 10
10 x 4 = 40
40 ÷ 2 = 20
20 x 4 = 80
the missing number is 40.

3. Evaluate 7P5 using formula

Solution: Formula for number of permutation = nPr = $\frac{n!}{\left ( n-r \right )!}$
when n= 7 and r = 5, assign the value in the given formula.

7P5 = $\frac{7!}{\left ( 7-5 \right )!}=\frac{7!}{(2)!}=\frac{7*6*5*4*3*2*1}{2*1}$ = 7*6*5*4*3 = 2520

4. Find the number of combinations of the letter of the word “KEYBOARD” when 5 letter is taken at a time.

Solution: The word “KEYBOARD” has 8   different letters, when 5 letter are taken at a time

The number of combination = ncr = $\frac{n!}{\left ( n-r \right )!r!}$
here n = 8 r = 5             nCr = $\frac{8!}{\left ( 8-4 \right )!4!}$
= $\frac{8!}{(4)!4!}=\frac{8*7*6*5*4*3*2*1}{(4*3*2*1)4*3*2*1}$ = 70

5. Find the value of n when the first term of the series is 10, their common difference is 3 and the nth term Tn = 64

Solution: Given the first term a = 10, d = 3, Tn = 64
The general form the series is Tn = a +(n - 1)d
assign the value of the term here
Tn = a + (n - 1)d
64 = 10 + (n-1)3
64 = 10 + 3n -3
64 = 7 + 3n   (add -7 on side of the equation)
64 - 7 = 7 + 3n - 7
57 = 3n
$\frac{57}{3}$ = n
19 = n

6. Find the sum of the series of geometric progression term 3+6+12+.......to 10 terms

Solution: Given a = 2,  n = 10
r = $\frac{T_{2}}{T_{1}}=\frac{6}{3}$ = 2

= $\frac{T_{3}}{T_{2}}=\frac{12}{6}$ = 2

since r is greater than 1 we have sum of series of term formula  Sn = $\frac{a\left ( r^{n}-1 \right )}{\left ( r-1 \right )}$

S10 = $\frac{2\left ( 2^{10}-1 \right )}{\left ( 2-1 \right )}$

S10 = $\frac{2\left ( 1024-1 \right )}{\left ( 1 \right )}$

S10 = 2 (1023)
S10 = 2046