Chapter 12.6, Problem 35E

(a)

To determine

To find:the number of possible groups of answers to the questions have exactly $8$ correct answers of false.

Answer to Problem 35E

The possible groups having exactly $8$ correct answer of false are $495$ .

Explanation of Solution

Given:

Rafael is taking test that contains a section of $12$ true false questions.

Concept used:

The binomial expansion is in the form of:

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Where here $x=\text{true},y=\text{false}$ .

Calculation:

Rafael is taking test that contains a section of true false questions.

Let $t$ represents the answer as true and $f$ represents the answer as false.

Find the number of possible groups.

For this expand:

  ${\left(t+f\right)}^{12}$ .

The expansion will have thirteen terms. Find the first seven terms using the sequence:

  $1,\frac{12}{1},\frac{12.11}{2!}or66,\frac{\mathrm{12.11.10}}{3!}or220,\frac{\mathrm{12.11.10.9}}{4!}or\mathrm{495......}\frac{\mathrm{12.11.10.9.8.7.6}}{7!}or792.$

Then use the symmetry to find the remaining terms:

  $\begin{array}{l}{\left(t+f\right)}^{12}=t^{12}+12t^{11}+66t^{10}{f}^2+220t^9{f}^3+495t^8{f}^4+792t^7{f}^5+\\ 924t^6{f}^6+792t^5{f}^7+495t^4{f}^8+220t^3{f}^9+66t^2{f}^{10}+12t{f}^{11}+f^{12}.\end{array}$

To find the possible group that’s have exactly $8$ correct answers of false, the total number of groups is the same as the coefficient of $t^4{f}^8$ this is $495$ .

Hence, the possible groups having exactly $8$ correct answer of false are $495$ .

(b)

To determine

To find:the number of possible groups of answers to those questions have at least $6$ correct answers of true.

Answer to Problem 35E

There are $2510$ possible groups of questions having at least $6$ correct answers of true.

Explanation of Solution

Given:

Rafael is taking test that contains a section of $12$ true false questions.

Concept used:

The binomial expansion is in the form of:

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Where here $x=\text{true},y=\text{false}$ .

Calculation:

Rafael is taking test that contains a section of true false questions.

Let $t$ represents the answer as true and $f$ represents the answer as false.

Find the number of possible groups.

For this expand:

  ${\left(t+f\right)}^{12}$ .

The expansion will have thirteen terms. Find the first seven terms using the sequence:

  $1,\frac{12}{1},\frac{12.11}{2!}or66,\frac{\mathrm{12.11.10}}{3!}or220,\frac{\mathrm{12.11.10.9}}{4!}or\mathrm{495......}\frac{\mathrm{12.11.10.9.8.7.6}}{7!}or792.$

Then use the symmetry to find the remaining terms:

  $\begin{array}{l}{\left(t+f\right)}^{12}=t^{12}+12t^{11}+66t^{10}{f}^2+220t^9{f}^3+495t^8{f}^4+792t^7{f}^5+\\ 924t^6{f}^6+792t^5{f}^7+495t^4{f}^8+220t^3{f}^9+66t^2{f}^{10}+12t{f}^{11}+f^{12}.\end{array}$

To have the at least $6$ correct answers of true means, there can be $6,7,8,9,10,11or12$ .

The total number of groups is same as the sum of the coefficients of

  $t^6{f}^6,t^7{f}^5,t^8{f}^4,t^9{f}^3,t^{10}{f}^2,t^{11}f\text{ or }{t}^{12}.$

This sum is:

  $1+12+66+220+495+792+924=2510.$

Hence, there are $2510$ possible groups of questions having at least $6$ correct answers of true.