The number of ways of arranging letters of the word havana so that v and n do not appear together is

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Hint: First we find the total number of arrangements of the letters of the word 'BANANA'.Then take two of the N's together i.e., take ‘NN’ as one block, and find the number of such permutations where the N’s appear together or adjacently. Now, note that the required number of permutations of the letters of the word 'BANANA' in which the two N’s do not appear adjacently is given by\[ = \]Total number of arrangements \[ - \]Number of arrangements where two N’s appear together

Complete step by step Answer:

There is a total of 6 letters in the word ‘BANANA’ out of which N repeats 2 times and A repeats 3 times.The total number of arrangements of the letters of the word 'BANANA' is \[\dfrac{{6!}}{{3!2!}}\; = \dfrac{{720}}{{6 \times 2}} = \dfrac{{720}}{{12}} = 60\] [divide by \[\left( {3!2!} \right)\]due to the repetitions of A and N]Now we write two of the N's together i.e. take ‘NN’ as one block, B A A A NN So now basically we have 5 letters in total.Therefore, Number of such permutations where the two N’s appear together or adjacently, is $\dfrac{{5!}}{{3!}} = \dfrac{{120}}{6} = 20$ [considering the repetitions of 3A′s]Hence, the required number of permutations of the letters of the word 'BANANA' in which the two N’s do not appear adjacently is given by\[ = \]Total number of arrangements \[ - \]Number of arrangements where two N’s appear together\[ = 60 - 20\] \[ = 40\]

The number of arrangements of the letters of the word 'BANANA' in which the two N’s do not appear adjacently is 40.

Note: A permutation is an act of arranging the objects or numbers in order while Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.

The formula for permutations is given by: \[{}^n{P_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{\left( {n - r} \right)!}}\] The formula for combinations is given by: \[{}^n{C_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]In the case of repetitions while arranging: Since in the word BANANA, there are 3 A’s and 2 N’s, therefore while calculating the number of arrangements one must consider the repetitions of the letters. Note that if there are n things to be arranged in a row, among which things are of one kind, b things are of another kind, and c things are of another, then the total number of arrangements is given by $\dfrac{{n!}}{{a!b!c!}}$.

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• Correct Answer: B

  Solution :

  Given word is HAVANA(3A, 1H, 1N, 1V) Total number of ways arranging the given word \[=\frac{6!}{3!}=120\] Total number of words in which N, V together \[=\frac{5!}{3!}\times 2!=40\] \[\therefore \] Required number of ways \[=120-40=80\]

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Page 2

• Correct Answer: A

  Solution :

  Let E be the event that a six occurs and A be the event that man reports that It is a six. \[\therefore \]\[P(E)=\frac{1}{6},P(E)=\frac{5}{6},P\left( \frac{A}{E} \right)=\frac{3}{4}\] and        \[P\left( \frac{A}{E} \right)=\frac{1}{4}\] Using Bays theorem                 \[P\left( \frac{E}{A} \right)=\frac{P(E).P\left( \frac{A}{E} \right)}{P(E).P\left( \frac{A}{E} \right)+P(E)P\left( \frac{A}{E} \right)}\]       \[=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}\]

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Page 3

• Correct Answer: B

  Solution :

  \[{{\cos }^{2}}(A-B)+oc{{s}^{2}}B-2\cos (A-B)cos\,A\,cos\,B\] \[={{\cos }^{2}}(A-B)+co{{s}^{2}}B-\cos (A-B)\] \[[cos(A+B)+cos(A-B)]\]                 \[={{\cos }^{2}}B-\cos (A-B)cos(A+B)\]                 \[={{\cos }^{2}}B-\frac{1}{2}[cos2A+cos2B]\] \[={{\cos }^{2}}B-\frac{1}{2}[2co{{s}^{2}}B-1+cos2A]\] \[=\frac{1}{2}-\frac{1}{2}\cos 2A\] \[\frac{1}{2}-\frac{1}{2}(2co{{s}^{2}}A-1)\] \[=1-{{\cos }^{2}}A={{\sin }^{2}}A\] Hence, it is independent of B.

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Page 4

• Correct Answer: A

  Solution :

  The intersection point of \[x+2y=3\]and \[3x+4y=7\]is \[(1,1).\] For consistent, point\[(1,1)\]satisfies the equation \[ax+y=3\]                 \[\therefore \]  \[a+1=3\]                 \[\Rightarrow \]               \[a=2\]

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Page 5

• Correct Answer: D

  Solution :

  Given, \[\left| \begin{matrix}    x-1 & 5x & 7  \\    {{x}^{2}}-1 & x-1 & 8  \\    2x & 3x & 0  \\ \end{matrix} \right|=a{{x}^{2}}+b{{x}^{2}}+cx+d\] LHS\[=(x-1)(0-24x)-5x(0-16x)\] \[+\,7(3{{x}^{2}}-3x-2{{x}^{2}}+2x)\] \[=-24{{x}^{2}}+24x+80{{x}^{2}}+21{{x}^{3}}-14{{x}^{2}}-7x\] \[=21{{x}^{3}}+42{{x}^{2}}+17x\] \[\therefore \]  \[21{{x}^{3}}+42{{x}^{2}}+17x\] \[=a{{x}^{3}}+b{{x}^{2}}+cx+d\]                 \[\Rightarrow \]               \[c=17\]                

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Page 6

• Correct Answer: B

  Solution :

  Using \[AM\ge GM\] \[\frac{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}{3}\ge \sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\]                                                                 ?(i) Again using \[AM\ge GM\] \[\frac{a+b}{2}\ge \sqrt{ab},\frac{b+c}{2}\ge \sqrt{bc}\frac{c+a}{2}\ge \sqrt{ca}\] \[\Rightarrow \]\[(a+b)(b+c)(c+a)\ge 8abc\] \[\Rightarrow \]\[\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}\le \frac{1}{2}\] \[\therefore \]From Eq. (i)          \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le \frac{3}{2}\]

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Page 7

• Correct Answer: C

  Solution :

  Given, \[\tan \,{{1}^{o}}\tan {{2}^{o}}...\tan {{89}^{o}}={{x}^{2}}-8\] \[\Rightarrow \]\[\tan {{1}^{o}}\tan {{2}^{o}}...\tan {{44}^{o}}\tan {{45}^{o}}\cot {{44}^{o}}\] \[\Rightarrow \]\[...\,\cot {{2}^{o}}\cot {{1}^{o}}={{x}^{2}}-8\] \[\Rightarrow \]\[1={{x}^{2}}-8\Rightarrow {{x}^{2}}=9\] \[\Rightarrow \]\[x=\pm 3\]

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Page 8

• Correct Answer: C

  Solution :

  Given \[\frac{\sin (x+3\alpha )}{\sin (\alpha -x)}=3\] Applying componendo and dividendo, we get  \[\Rightarrow \]\[\frac{\sin (x+3\alpha )+\sin (\alpha -x)}{\sin (x+3\alpha )-\sin (\alpha -x)}=\frac{3+1}{3-1}\] \[\Rightarrow \]\[\frac{2\sin 2\alpha \cos (\alpha +x)}{2\cos 2\alpha \sin (\alpha +x)}=2\] \[\Rightarrow \]\[\frac{\tan 2\alpha }{\tan (\alpha +x)}=2\] \[\Rightarrow \]\[\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\times \frac{(1-\tan \alpha \tan x)}{(\tan \alpha +\tan x)}=2\] \[\Rightarrow \]\[\tan \alpha -{{\tan }^{2}}\alpha \tan x=\tan \alpha +\tan x\]                 \[-{{\tan }^{3}}\alpha -{{\tan }^{2}}\alpha \tan x\] \[\Rightarrow \]\[\tan x={{\tan }^{3}}\alpha \]

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Page 9

• Correct Answer: D

  Solution :

  \[2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac\] \[\Rightarrow \]\[{{a}^{2}}+{{(2b)}^{2}}-4ab+{{a}^{2}}+{{c}^{2}}-2ac=0\] \[\Rightarrow \]\[{{(a-2b)}^{2}}+{{(a-c)}^{2}}=0\] \[\Rightarrow \]\[a=2b=c\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[=\frac{{{c}^{2}}+{{c}^{2}}-{{\left( \frac{c}{2} \right)}^{2}}}{2\times c\times c}=\frac{2{{c}^{2}}-\frac{{{c}^{2}}}{4}}{2{{c}^{2}}}\] \[\Rightarrow \]               \[\cos B=\frac{7}{8}\]

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Page 10

• Correct Answer: C

  Solution :

  Since, \[P(A\cup B)=P(A)+P(B)-P(A)P(B)\] \[[\because \,P(A\cap B)=P(A)P(B)]\]                 \[\Rightarrow \]\[0.8=0.3+P(B)-0.3P(B)\]                 \[\Rightarrow \]\[0.7P(B)=0.5\Rightarrow P(B)=\frac{5}{7}\]

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Page 11

• Correct Answer: D

  Solution :

  \[\therefore \]Required probability \[=\frac{{{\,}^{4}}{{C}_{2}}+{{\,}^{5}}{{C}_{2}}}{{{\,}^{9}}{{C}_{2}}}\]                              \[=\frac{6+10}{36}=\frac{16}{36}\] \[=\frac{4}{9}\]

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Page 12

• Correct Answer: B

  Solution :

  \[\frac{1}{\cos {{290}^{o}}}+\frac{1}{\sqrt{3}\sin {{250}^{o}}}\] \[=\frac{1}{\cos {{70}^{o}}}-\frac{1}{\sqrt{3}\sin {{110}^{o}}}\] \[=\frac{\sqrt{3}\sin {{10}^{o}}-\cos {{70}^{o}}}{\sqrt{3}\sin {{110}^{o}}\cos {{70}^{o}}}\] \[=\frac{\sqrt{3}\sin ({{180}^{o}}-{{70}^{o}})-cos{{70}^{o}}}{\sqrt{3}\sin (180-{{70}^{o}})cos{{70}^{o}}}\] \[=\frac{\frac{\sqrt{3}}{2}\sin {{70}^{o}}-\frac{1}{2}\cos {{70}^{o}}}{\frac{\sqrt{3}}{2}\sin 70\cos {{70}^{o}}}\] \[\frac{\cos {{30}^{o}}\sin {{70}^{o}}-\sin {{30}^{o}}\cos {{70}^{o}}}{\frac{\sqrt{3}}{2}.\frac{1}{2}\sin {{140}^{o}}}\] \[=\frac{\sin ({{70}^{o}}-{{30}^{o}})}{\frac{\sqrt{3}}{4}\sin ({{180}^{o}}-{{40}^{o}})}\] \[=\frac{\sin {{40}^{o}}}{\frac{\sqrt{3}}{4}\sin {{40}^{o}}}=\frac{4}{\sqrt{3}}\]

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Page 13

• Correct Answer: A

  Solution :

  Equation of any line through (1, 2) is \[y-2=m(x-1)\]                 \[\Rightarrow \]               \[y-mx+m-2=0\] The distance of line \[y=2x\]from the point (3, 1) is greatest. \[\therefore \]Required line in \[y=2x.\]

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Page 14

• Correct Answer: B

  Solution :

  Let \[{{S}_{1}}={{x}^{2}}+{{y}^{2}}=9\] \[PA.PB={{(\sqrt{{{S}_{1}}})}^{2}}\] \[={{\left( \sqrt{{{(3)}^{3}}+{{(11)}^{2}}-9} \right)}^{2}}=121\]

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Page 15

• Correct Answer: C

  Solution :

  \[{{x}^{2}}+{{y}^{2}}+6x+6y-2=0\] Centre \[(-3,-3),\]radius \[=\sqrt{9+9+2}\]                 \[=\sqrt{20}\]
  Now, \[QC=\sqrt{{{(-3)}^{2}}+{{6}^{2}}}=\sqrt{45}\] In right \[\Delta CPQ\] \[PQ=\sqrt{45-20}=5\]

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Page 16

• Correct Answer: C

  Solution :

  Let sides of the triangle are \[4x,5x,6x.\] \[s=\frac{4x+5x+6x}{2}=\frac{15}{2}x\] \[\Delta =\sqrt{\frac{15}{2}x\left( \frac{15}{2}x-4x \right)\left( \frac{15}{2}x-5x \right)\left( \frac{15}{2}x-6x \right)}\] \[=\sqrt{\frac{15}{2}x\times \frac{7}{2}x\times \frac{5}{2}x\times \frac{3}{2}x}\] \[=\frac{15\sqrt{7}{{x}^{2}}}{4}\]                 Circumradius,  \[R=\frac{4x\times 5x\times 6x}{4\times \frac{15\sqrt{7}{{x}^{2}}}{4}}\]                                                 \[=\frac{8}{\sqrt{7}}x\]                 Inradius, \[r=\frac{\frac{15\sqrt{7}}{4}{{x}^{2}}}{\frac{15}{2}x}\]                                 \[=\frac{\sqrt{7}}{2}xax\]                                 \[\frac{R}{r}=\frac{\frac{8x}{\sqrt{7}}}{\frac{\sqrt{7x}}{2}}=\frac{16}{7}\]             

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Page 17

• Correct Answer: C

  Solution :

  \[4{{\sin }^{-1}}x+{{\cos }^{-1}}x=\pi \] \[\Rightarrow \]\[4{{\sin }^{-1}}x+\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)=\pi \] \[\Rightarrow \]\[3{{\sin }^{-1}}x=\pi -\frac{\pi }{2}=\frac{\pi }{2}\] \[\Rightarrow \]\[{{\sin }^{-1}}x=\frac{\pi }{6}\] \[\Rightarrow \]\[x=\sin \frac{\pi }{6}=\frac{1}{2}\]

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Page 18

• Correct Answer: A

  Solution :

  \[\cos \left( \frac{1}{2}{{\cos }^{-1}}\frac{1}{8} \right)=x\]            (let) \[\Rightarrow \]\[{{\cos }^{-1}}\frac{1}{8}=2{{\cos }^{-1}}x\] \[={{\cos }^{-1}}(2{{x}^{2}}-1)\] \[\Rightarrow \]\[\frac{1}{8}=2{{x}^{2}}-1\] \[\Rightarrow \]\[2{{x}^{2}}=\frac{1}{8}+1=\frac{9}{8}\] \[\Rightarrow \]\[{{x}^{2}}=\frac{9}{16}\] \[\Rightarrow \]\[x=\sqrt{\frac{9}{16}}=\frac{3}{4}\]       \[[\because \,0\le x\le 1]\]

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Page 19

• Correct Answer: D

  Solution :

  Given vectors are coplanar, if                 \[\left| \begin{matrix}    1 & 1 & m  \\    1 & 1 & m+1  \\    1 & -1 & m  \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix}    0 & 0 & -1  \\    1 & 1 & m+1  \\    1 & -1 & m  \\ \end{matrix} \right|=0\]                              \[[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}]\] \[\Rightarrow \]\[-1(-1-1)=0\] \[\Rightarrow \]\[2\ne 0\] \[\therefore \]No value of m for which vectors are coplanar.

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Page 20

• Correct Answer: D

  Solution :

  Given, \[\vec{a}+\vec{b}+\vec{c}=0\] \[\Rightarrow \]\[{{(\vec{a}+\vec{b}+\vec{c})}^{2}}=0\] \[\Rightarrow \]\[|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}\] \[+\,2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\]                 \[\Rightarrow \]\[4+9+16+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\] \[\Rightarrow \]\[\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=\frac{-29}{2}\]

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Page 21

• Correct Answer: C

  Solution :

  Here \[a=2,m=-1\] \[\therefore \]Required point is \[(a{{m}^{2}},-2am)=(2,4)\]

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Page 22

• Correct Answer: C

  Solution :

  Given curve is \[16{{x}^{2}}+25{{y}^{2}}=400\] \[\Rightarrow \]\[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\] Here \[a=5,b=4\] \[{{F}_{1}}\]and \[{{F}_{2}}\]are forcus. \[\therefore \]\[P{{F}_{1}}+P{{F}_{2}}=2a=10\]

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Page 23

• Correct Answer: C

  Solution :

  Let the hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Eccentricity, \[e=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}\] \[\Rightarrow \]               \[\frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{({{a}^{2}}+{{b}^{2}})}\] Conjugate hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1\] Eccentricity, \[{{e}_{1}}=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}}\] \[\Rightarrow \]               \[\frac{1}{e_{1}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] Now, \[\frac{1}{{{e}^{2}}}+\frac{1}{e_{1}^{2}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\]

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Page 24

• Correct Answer: C

  Solution :

  \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{3}}x}{x\sin x\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2(1-{{\cos }^{3}}x)}{x\sin 2x}\] [Using LHospital Rule] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2[-3{{\cos }^{2}}x(-\sin x)]}{\sin 2x+x+\cos 2x.2}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6{{\cos }^{2}}x\sin x}{\sin 2x+2x\cos 2x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6[-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x]}{2\cos 2x+2[-x\sin 2x.2+\cos 2x]}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6[-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x]}{2\cos 2x-4x\sin 2x+2\cos 2x}\] \[=\frac{6}{2+2}\] \[=\frac{3}{2}\]

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Page 25

• Correct Answer: A

  Solution :

  \[y={{\cos }^{-1}}(\cos x)\] \[y=\frac{-1}{\sqrt{1-{{\cos }^{2}}x}}(-\sin x)\] \[=\frac{\sin x}{\sqrt{{{\sin }^{2}}x}}\] \[=1\forall x\]

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Page 26

• Correct Answer: B

  Solution :

  Position vector of median \[\overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}\] \[=\frac{(-3+5)\hat{i}(0-2)\hat{j}+(4+4)\hat{k}}{2}\] \[=\hat{i}-\hat{j}+4\hat{k}\] \[\therefore \]\[|\overrightarrow{AD}|=\sqrt{{{1}^{2}}+{{(-1)}^{2}}+{{(4)}^{2}}}=\sqrt{18}\]

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