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Q20E

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Discrete Mathematics and its Applications
Found in: Page 413
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

How many bit strings of length \({\bf{10}}\) have

a) exactly three \(0s\)?

b) more \(0s\) than \(1s\) ?

c) at least seven \(1s\) ?

d) at least three \(1s\) ?

a) The bit string of length \(10\) having exactly three \(0's\)is \(120\).

b) The bit string of length \(10\) having more \(0's\)than \(1's\)is \(386\).

c) The bit string of length \(10\) having at least seven \(1's\)is \(176\).

d) The bit string of length \(10\) having at least three \(1's\)is \(968\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given data

Length of bit string \( = 10\)

Step 2: Concept of Combination

A combination is a selection of items from a set that has distinct members.

Formula:

\(_n{C_r} = \frac{{n!}}{{r!(n - r)!}}\)

Step 3: Calculation for the bit string for exactly three \(0's\)

a)

Find the bit string of length \(10\) having exactly three \(0's\), use the formula of combination.

Number of possible outcomes \( = C(10,3)\)

\(\begin{array}{l}C(10,3) = \frac{{10!}}{{3!(10 - 3)!}}\\C(10,3) = \frac{{10!}}{{3!7!}}\end{array}\)

Simplify the factorial of the above expression:

\(\begin{array}{l}C(10,3) = \frac{{10 \times 9 \times 8 \times 7!}}{{3 \times 2 \times 1 \times 7!}}\\C(10,3) = \frac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}}\\C(10,3) = 120\end{array}\)

Hence, the bit string of length 10 having exactly three \(0's\)is \(120\) .

Step 4: Calculation for the bit string for more \(0's\)than \(1's\)

b)

Find the bit string of length \(10\) having more \(0's\) than \(1's\), when less than\(5\)bits are \(0's\), use the formula of combination.

Number of possible outcomes \(\; = C(10,4) + C(10,3) + C(10,2) + C(10,1)\)

\(\begin{array}{l}C(10,4) + C(10,3) + C(10,2) + C(10,1) + C(10,0) = \left( {\frac{{10!}}{{4!(10 - 4)!}} + \frac{{10!}}{{3!(10 - 3)!}} + \frac{{10!}}{{2!(10 - 2)!}} + \frac{{10!}}{{1!(10 - 1)!}} + \frac{{10!}}{{0!(10 - 0)!}}} \right)\\C(10,4) + C(10,3) + C(10,2) + C(10,1) + C(10,0) = \frac{{10!}}{{4!6!}} + \frac{{10!}}{{3!7!}} + \frac{{10!}}{{2!8!}} + \frac{{10!}}{{1!9!}} + \frac{{10!}}{{0!10!}}\end{array}\)

Simplify the factorial of the above expression:

\(\begin{array}{l}C(10,4) + C(10,3) + C(10,2) + C(10,1) + C(10,0) = 210 + 120 + 45 + 10 + 1\\C(10,4) + C(10,3) + C(10,2) + C(10,1) + C(10,0) = 386\end{array}\)

Hence, the bit string of length 10 having more \(0's\) than \(1's\)is \(386\).

Step 5: Calculation for the bit string for at least seven \(1's\]

c)

Find the bit string of length \(10\] having at least seven \(1's\], use the formula of combination.

The bit string of length 10 having at least seven \(1's\] is same as the bit string of length \(10\] having at most three \(0's\].

Number of possible outcomes \( = C(10,3) + C(10,2) + C(10,1) + C(10,0)\]

\(\begin{array}{l}C(10,3) + C(10,2) + C(10,1) + C(10,0) = \frac{{10!}}{{3!(10 - 3)!}} + \frac{{10!}}{{2!(10 - 2)!}} + \frac{{10!}}{{1!(10 - 1)!}} + \frac{{10!}}{{0!(10 - 0)!}}\\C(10,3) + C(10,2) + C(10,1) + C(10,0) = \frac{{10!}}{{3!7!}} + \frac{{10!}}{{2!8!}} + \frac{{10!}}{{1!9!}} + \frac{{10!}}{{0!10!}}\end{array}\]

Simplify the factorial of the above expression:

\(\begin{array}{l}C(10,3) + C(10,2) + C(10,1) + C(10,0) = 120 + 45 + 10 + 1\\C(10,3) + C(10,2) + C(10,1) + C(10,0) = 176\end{array}\]

Hence, the bit string of length 10 having at least seven \(1's\] is \(176\].

Step 6: Calculation for the bit string for at least three \(1's\]

d)

Find the bit string of length \(10\] having at least three \(1's\], use the formula of combination.

Number of possible outcomes\( = C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)\]

\(\begin{array}{l}C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10) = \left( {\begin{array}{*{20}{l}}{\frac{{10!}}{{3!(10 - 3)!}} + \frac{{10!}}{{4!(10 - 4)!}} + \frac{{10!}}{{5!(10 - 5)!}} + \frac{{10!}}{{6!(10 - 6)!}}}\\{ + \frac{{10!}}{{7!(10 - 7)!}} + \frac{{10!}}{{8!(10 - 8)!}} + \frac{{10!}}{{9!(10 - 9)!}} + \frac{{10!}}{{10!(10 - 10)!}}}\end{array}} \right]\\C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10) = \frac{{10!}}{{3!7!}} + \frac{{10!}}{{4!6!}} + \frac{{10!}}{{5!5!}} + \frac{{10!}}{{6!4!}} + \frac{{10!}}{{7!3!}} + \frac{{10!}}{{8!2!}} + \frac{{10!}}{{9!1!}} + \frac{{10!}}{{10!0!}}\end{array}\]

Simplify the factorial of the above expression:

\(\begin{array}{l}C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10) = 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1\\C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10) = 968\end{array}\]

Hence, the bit string of length 10 having at least three \(1's\]is \(968\].

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