mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm

### <Success>Problems in combination</Success>
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<h3 id="successexample-1success"><Success>Example 1</Success></h3>
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<p><strong>(JEE2014)</strong>   Let $n \geqslant 2$ be an integer. Take $n$ distinct points on a circle and join each pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then find the value of $n$.</p>
<p><strong>Solution:</strong>  These are $n$ line segments joining adjacent points.
So the number of blue line
segments $=n$.</p>
<p>If we join all the points, the total number of line segments $^n{c_2}=\frac{n(n-1)}{2}$</p>
<p>So the number of line segment which are coloured red is : $\frac{n(n-1)}{2}-n$</p>
<p>Now $\frac{n(n-1)}{2}-n=n$
$$
\Rightarrow n(n-1)=4 n \Rightarrow n^2=5 n \text {. So } n=5 \text {. }
$$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1390a5c8-579c-48d7-0e42-1090fb4e3b00/public"></p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-3.jpg"><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-3a.jpg"></p>
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### <Success>Solution </Success>
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<p><strong>(iv)</strong> In how many words letters $A, E, O$ occur only in odd positions?</p>
<p>There are $3 E^{\prime}$ s. Total odd positions are $1,3,5,7,9$. So in 5 places, $A, O$ and $3 E^{\prime} s$ can be placed in $\frac{5 !}{3 !}$ ways.</p>
<p>In remaining 4 positions $L, D \ \text{and} \ 2 \mathrm{~N}$ ’s can be placed in $\frac{4 !}{2 !}$ ways.</p>
<p>So total number of such words $=\frac{5 !}{3 !} \times \frac{4 !}{2 !}=240$</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-7.jpg">
<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-7a.jpg">
<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-7b.jpg"></p>
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### <Success>Solution</Success>
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<p>(ii) 4 girls who stand consecutively, can be treated as one unit. So 5 boys and these 4 girls can stand in 6! ways.</p>
<p>Further these 4 girls can be permuted in 4 ! ways. The girl who stands separately can be choosen from 5 girls in ${ }^5 C_1$ ways.</p>
<p>The last girl can be placed in 5 ways (as it can be placed adjacent to previous 4 girls). So now by multiplication principle, the total number of arrangements</p>
<p>$=6 ! \times 4 ! \times{ }^5 C_9 \times 5=432000$</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-9.jpg">
<img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/284d7d94-af5a-4ffc-8544-0bbdc72b9100/public">
<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-9a.jpg"></p>
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### <Success>Example 7</Success>
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<p>Lisa chooses 13 cards from a deck of 52 cards. In how many ways she can choose so that she gets 2 kings & 2 queens.</p>
<p><strong>Solution:</strong></p>
<p>${ }^4 c_2 \times{ }^4 c_2 \times{ }^{44} c_9$</p>
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<p>======</p>
<h3 id="successexample-8success"><Success>Example 8</Success></h3>
<p>In how many ways 3 balls can be choosen from a set of 3 red, 4 blue and 2 green balls so that</p>
<p>(i) all are of different colours?</p>
<p>ii) all are of the same colour?</p>
<p><strong>Solution:</strong></p>
<p>$\text{(i)} \ \binom{3}{1} \binom{4}{1}  \binom{2}{1}=24$</p>
<p>$\text{(ii)} \ \binom{3}{3} \binom{4}{3} = 1 + 4 =5$</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-10.jpg"></p>
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### <Success>Example 9</Success>
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<p>All permutations of the word ’ $L A S T ‘$ are arranged according to english dictionary ordering. what is the positionss of the word ’ $S A L T$ ’ in this ordering?</p>
<p><strong>Solution:</strong>   The number of words starting with</p>
<p>$$
\text { A } \square \square \square \rightarrow 3 !=6
$$</p>
<p>The number of words starting with</p>
<p>$$
\text { L } \square \square \square \rightarrow 3 !=6
$$</p>
<p>Next one will be $L A S T^{\prime}$ $\rightarrow 13^{\text {th }}$ positions</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-08-chapter-01-permutation-combination-L-1-5-vajjv1t6rnm-11.jpg"></p>
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<p>======</p>
<h3 id="successthank-yousuccess"><Success>Thank you</Success></h3>
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