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Q20E

Expert-verifiedFound in: Page 413

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**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 \(968\).

Length of bit string \( = 10\)

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

**Formula:**

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

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 \(120\) .

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 \(386\).

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\].

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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