Explanation
$${\textbf{Step 1}}\;:\;{\mathbf{Finding}}\;{\mathbf{number}}\;{\mathbf{of}}\;{\mathbf{words}}\;{\mathbf{starting}}\;{\mathbf{with}}\;{\mathbf{the}}\;{\mathbf{alphabets}}\;{\mathbf{in}}\;{\mathbf{the}}\;{\mathbf{given}}\;{\mathbf{word}}$$
$${\text{No}}.{\text{ of words beginning with A }} = 4! = 4 \times 3 \times 2 = 24$$
$${\text{No}}.{\text{ of words beginning with N }} = 4! = 4 \times 3 \times 2 = 24$$
$${\text{No}}.{\text{ of words beginning with R }} = 4! = 4 \times 3 \times 2 = 24$$
$${\text{No}}.{\text{ of words beginning with U }} = 4! = 4 \times 3 \times 2 = 24$$
$${\text{So}},{\text{ in total }}96{\text{ words will be formed while beginning with letter A}},{\text{ N}},{\text{ R and U}}.$$
$${\textbf{Step 2}}\;\;:\;{\mathbf{Finding}}\;{\mathbf{the}}\;{\mathbf{order}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{word}}$$
$${\text{Order of }} 97{\text { th word }} - {\text{ VANRU}}$$
$${\text{Order of }}98{\text{th word }} - {\text{ VANUR}}$$
$${\text{Order of }}99{\text{th word }} - {\text{ VARNU}}$$
$${\text{Order of }}100{\text{th word}} - {\text{ VARUN}}.$$
$${\mathbf{Therefore}},\;{\mathbf{rank}}\;{\mathbf{of}}\;{\mathbf{word}}\;'{\mathbf{VARUN}}'\;{\textbf{is 100. Option C is correct.}}$$
$${\textbf{Step -1: Simplifying the problem.}}$$
$${}^{13}{C_2} + {}^{13}{C_3} + ...... + {}^{13}{C_{13}}.$$
$$ = {}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + {}^{13}{C_3} + ...... + {}^{13}{C_{13}} - {}^{13}{C_0} - {}^{13}{C_1}.$$
$$ = \left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right) - \left( {{}^{13}{C_0} + {}^{13}{C_1}} \right){\text{ }} \to \left( {\text{i}} \right).$$
$${\textbf{Step -2: Deduce the value of }}\mathbf{\left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right)}{\textbf{ using summation formula}}{\textbf{. }}$$
$$\sum\limits_{i = 0}^n {{}^n{C_i}} = {2^n}{\text{ }} \to \left( {{\text{ii}}} \right)$$
$$ \Rightarrow \left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right) = \sum\limits_{i = 0}^{13} {{}^{13}{C_i}.} $$
$$ \Rightarrow \sum\limits_{i = 0}^{13} {{}^{13}{C_i} = {2^{13}}({\text{By using (}}ii{\text{)}})} .$$
$${\text{then, using (i) and (ii) we get}},$$
$$\therefore {2^{13}} - \left( {{}^{13}{C_0} + {}^{13}{C_1}} \right).$$
$$ = {2^{13}} - \left( {\dfrac{{13!}}{{13!0!}} + \dfrac{{13!}}{{12!1!}}} \right).$$
$$ = {2^{13}} - \left( {1 + 13} \right).$$
$$ = {2^{13}} - 14.$$
$${\textbf{Hence , the correct answer is (B) }}\mathbf{{2^{13}} - 14.}$$
We have,
A box contains pair of shoes $$=5$$
Selected pair of shoes $$=4$$
Then,
Exactly one pair of shoes obtained is
$$ {{=}^{5}}{{P}_{4}} $$
$$ =\dfrac{5!}{\left( 5-4 \right)!} $$
$$ =\dfrac{5!}{1!}=5! $$
$$ =5\times 4\times 3\times 2\times 1 $$
$$ =120 $$
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