37. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

37. Since out of 8 total questions at least 3 questions has to be attempted from each of part I and II containing 5 and 7 questions respectively we can have the choices.

(a) 3 questions from I and 5 questions from II selected in ⁵C₃×⁷C₅ ways.

(b) 4 questions from I and 4 questions from II selected in ⁵C₄×⁷C₄ ways.

(c) 5 questions from I and 3 questions from II selected in ⁵C₅×⁷C₃ ways.

Therefore, the required number of ways.

= (⁵C₃×⁷C₅) + (⁵C₄×⁷C₄) + (⁵C₅×⁷C₃)

= $\frac{5!}{3!\left(5-3\right)!}$ × $\frac{7!}{5!\left(7-5\right)!}$ + $\frac{5!}{4!\left(5-4\right)!}$ × $\frac{7!}{4!\left(7-4\right)!}$ + $\frac{5!}{5!\left(5-5\right)!}$ × $\frac{7!}{3!\left(7-3\right)!}$

= (10 × 21) + (5 × 35) + 35

= 210 + 175 + 35

...more

37. Since out of 8 total questions at least 3 questions has to be attempted from each of part I and II containing 5 and 7 questions respectively we can have the choices.

(a) 3 questions from I and 5 questions from II selected in ⁵C₃×⁷C₅ ways.

(b) 4 questions from I and 4 questions from II selected in ⁵C₄×⁷C₄ ways.

(c) 5 questions from I and 3 questions from II selected in ⁵C₅×⁷C₃ ways.

Therefore, the required number of ways.

= (⁵C₃×⁷C₅) + (⁵C₄×⁷C₄) + (⁵C₅×⁷C₃)

= $\frac{5!}{3!\left(5-3\right)!}$ × $\frac{7!}{5!\left(7-5\right)!}$ + $\frac{5!}{4!\left(5-4\right)!}$ × $\frac{7!}{4!\left(7-4\right)!}$ + $\frac{5!}{5!\left(5-5\right)!}$ × $\frac{7!}{3!\left(7-3\right)!}$

= (10 × 21) + (5 × 35) + 35

= 210 + 175 + 35

= 420

less

<p><strong>37.</strong> Since out of 8 total questions at least 3 questions has to be attempted from each of part I and II containing 5 and 7 questions respectively we can have the choices.</p><p> (a) 3 questions from I and 5 questions from II selected in ⁵C₃×⁷C₅ ways.</p><p> (b) 4 questions from I and 4 questions from II selected in ⁵C₄×⁷C₄ ways.</p><p> (c) 5 questions from I and 3 questions from II selected in ⁵C₅×⁷C₃ ways.</p><p>Therefore, the required number of ways.</p><p>= (⁵C₃×⁷C₅) + (⁵C₄×⁷C₄) + (⁵C₅×⁷C₃)</p><p>=&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>5</mn><mo>!</mo></mrow><mrow><mn>3</mn><mo>!</mo><mrow><mo> (</mo><mrow><mn>5</mn><mo>−</mo><mn>3</mn></mrow><mo>)</mo></mrow><mo>!</mo></mrow></mfrac></mrow></math></span>&nbsp;×&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>7</mn><mo>!</mo></mrow><mrow><mn>5</mn><mo>!</mo><mrow><mo> (</mo><mrow><mn>7</mn><mo>−</mo><mn>5</mn></mrow><mo>)</mo></mrow><mo>!</mo></mrow></mfrac></mrow></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>5</mn><mo>!</mo></mrow><mrow><mn>4</mn><mo>!</mo><mrow><mo> (</mo><mrow><mn>5</mn><mo>−</mo><mn>4</mn></mrow><mo>)</mo></mrow><mo>!</mo></mrow></mfrac></mrow></math></span>&nbsp;×&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>7</mn><mo>!</mo></mrow><mrow><mn>4</mn><mo>!</mo><mrow><mo> (</mo><mrow><mn>7</mn><mo>−</mo><mn>4</mn></mrow><mo>)</mo></mrow><mo>!</mo></mrow></mfrac></mrow></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>5</mn><mo>!</mo></mrow><mrow><mn>5</mn><mo>!</mo><mrow><mo> (</mo><mrow><mn>5</mn><mo>−</mo><mn>5</mn></mrow><mo>)</mo></mrow><mo>!</mo></mrow></mfrac></mrow></math></span>&nbsp;×&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>7</mn><mo>!</mo></mrow><mrow><mn>3</mn><mo>!</mo><mrow><mo> (</mo><mrow><mn>7</mn><mo>−</mo><mn>3</mn></mrow><mo>)</mo></mrow><mo>!</mo></mrow></mfrac></mrow></math></span></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1747823344phpAsXD4t.jpeg" alt="" width="524" height="103" loading="lazy"></picture></div></div><p>= (10 × 21) + (5 × 35) + 35</p><p>= 210 + 175 + 35</p><p>= 420</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment