• #Definitions
• #Q1 — Words from DELHI
• #Q2 — Words from ALLAHABAD
• #Q3 — Words from AALLBBCOEFGG
• #Q4 — LUCKNOW, vowels always together
• #Q5 — SUCCESS, vowels always together
• #Q6 — LUCKNOW, vowels never together
• #Q7 — LUCKNOW, vowels occupy even places
• #Q8 — SUCCESS, vowels occupy even places
• #Q9 — Numbers between 99 and 1000
• #Q10 — 6-digit numbers from 2,3,4,5,6,8
• #Q11 — Handshakes among 10 people
• #Q12 — Handshakes = 66, find no of people
• #Q13 — Diagonals of a 10-point convex polygon
• #Q14 — Team of 11 from 17 players with 2 bowlers
• #Q15 — Committee of 3 from 5 males and 4 females
• #Q16 — Vegetables MCQ
• #Q17 — Handshakes in a group of 11 MCQ
• #Q18 — 3-letter words, consonant condition MCQ
• #Q19 — Diagonals of a dodecagon MCQ
• #Q20 — Letters among 20 friends MCQ
• #Q21 — Arrangements of UGCCSIR MCQ
• #Revisited — Numbers between 99 and 1000 again

Maths | PDF

/Permutation and combination — Sunday, 18 April 2021, 11:38 PM/

\[ \text{Permutation} \;\longrightarrow\; \text{arrangements} \;\longrightarrow\; {}^{n}P_{r} \quad \text{permutation} \] \[ \text{Combination} \;\longrightarrow\; \text{selection} \;\longrightarrow\; {}^{n}C_{r} \quad \text{combination} \]

(Q.1) No of words formed by word "DELHI" (positions 1 2 3 4 5)

without repetition

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/5,1/4,2/3,3/2,4/1} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 5! = 5\times4\times3\times2\times1 = 120 \]

with repetition :-

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/5,1/5,2/5,3/5,4/5} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ \text{No of ways} \;=\; 5\times5\times5\times5\times5 \;=\; 5^5 \]

(Q.2) Do same if word is "ALLAHABAD" (positions 1 2 3 4 5 6 7 8 9)

\[ \text{No of arrangements} \;=\; \frac{9!}{4!\,2!} \]

(the \(4!\) is the no of times A repeated; the \(2!\) is the no of times 'L' is repeated)

(Q.3) Do same if word is "AALLBBCOEFGG" (positions 1 2 3 4 5 6 7 8 9 10 11 12)

Assume all letters are different \(= 12!\)

But A is repeated 2 times (No of ways of arranging 2) \(= 2!\)

L \(\to\) repeated 2 \(\to\) \(= 2!\)

B \(\to\) repeated 2 \(\to\) \(= 2!\)

G \(\to\) repeated 2 \(\to\) \(= 2!\)

So Correct no of ways

\[ = \frac{12!}{2!\,2!\,2!\,2!} \]

(4) How many words can be formed by using all the letters of the word "LUCKNOW" so that the vowels always come together?

(separating vowels & assuming these a single letter)

"LUCKNOW" \(\uparrow\) \(\uparrow\) (the vowels are U and O)

\begin{tikzpicture}[scale=1.0]
  \draw (0,0) rectangle (1.6,0.9);
  \node at (0.8,0.45) {$OU$};
  \foreach \i/\txt [count=\c from 1] in {1.6/L,2.6/C,3.6/K,4.6/N,5.6/W} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
  \node[below] at (0.8,-0.15) {\small 1};
  \node[below] at (2.1,-0.15) {\small 2};
  \node[below] at (3.1,-0.15) {\small 3};
  \node[below] at (4.1,-0.15) {\small 4};
  \node[below] at (5.1,-0.15) {\small 5};
  \node[below] at (6.1,-0.15) {\small 6};
\end{tikzpicture}

No of ways of arranging 6 obj. \(= 6! \times 2!\)

(\(2!\) is the No of ways of arranging the 2 objects, i.e. the 2 vowels inside their block)

Total arrangements \(= 6! \times 2!\)

(5) Do same is word is "SUCCESS"

\(S\,U\,C\,C\,E\,S\,S\) (vowels underlined: U, E)

\begin{tikzpicture}[scale=1.0]
  \draw (0,0) rectangle (1.6,0.9);
  \node at (0.8,0.45) {$UE$};
  \foreach \i/\txt in {1.6/S,2.6/C,3.6/C,4.6/S,5.6/S} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
  \node[below] at (0.8,-0.15) {\small 1};
  \node[below] at (2.1,-0.15) {\small 2};
  \node[below] at (3.1,-0.15) {\small 3};
  \node[below] at (4.1,-0.15) {\small 4};
  \node[below] at (5.1,-0.15) {\small 5};
  \node[below] at (6.1,-0.15) {\small 6};
\end{tikzpicture}

\[ \text{No of ways} \;=\; \left( \frac{6!}{2!\,3!} \right) \times 2! \]

('C' is repeated 2 times; 'S' is repeated 3 times; the \(\times 2!\) arranges the vowels — No of ways of arranging vowels)

(6) How many words can be formed by using all the letters of the word "LUCKNOW" so that the vowels never come together?

"LUCKNOW" (positions 1 2 3 4 5 6 7)

No of ways vowels never come together

\[ = \text{Total arrangements} \;-\; \text{Arrangements in which vowels come together} \] \[ = 7! \;-\; 6!\times2! \] \[ = 7\times6\times120 \;-\; 6\times120\times2 \;=\; 5\times6\times120 \] \[ = 3600 \]

(7) How many words can be formed by using all the letters of the word "LUCKNOW" so that the vowels occupy even places?

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/1,1/2,2/3,3/4,4/5,5/6,6/7} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
  \foreach \i in {1,3,5} {
    \draw[->] (\i+0.5,-0.6) -- (\i+0.5,-0.05);
  }
  \node[below] at (3.5,-0.7) {\small even places};
\end{tikzpicture}

Arranging 3 places for 2 vowels.

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/3,1/2} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 3\times2 = {}^{3}P_{2} = 6 \text{ ways.} \]

(No of places available for vowel 1; No of places available for vowel 2)

Once vowels have been placed we have 5 places left for 5 letters (LCKNW).

\[ \text{No of ways} \;=\; 6\times5! \;=\; 6! \]

(8) How many words can be formed by using all the letters of the word "SUCCESS" so that the vowels occupy even places?

\(S\,U\,C\,C\,E\,S\,S\) (vowels: U, E)

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/1,1/2,2/3,3/4,4/5,5/6,6/7} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
  \foreach \i in {1,3,5} {
    \draw[->] (\i+0.5,-0.6) -- (\i+0.5,-0.05);
  }
  \node[below] at (3.5,-0.7) {\small even places (2 vowels)};
\end{tikzpicture}

\[ = \text{No of ways vowels can occupy even places} \;\times\; \text{Total no of ways of arranging remaining letters} \] \[ = {}^{3}P_{2} \times \frac{5!}{3!\,2!} \] \[ = \frac{3!}{1!} \times \frac{5!}{3!\,2!} \;=\; 5\times4\times3 \;=\; \underline{60} \]

Ques: How many numbers between 99 and 1000 can be formed with the digits \(0,1,2,3,4,5\)

We need to form 3-digit numbers.

Total no of objects to be arranged \(= 0,1,2,3,4,5 = 6\) (we can not have '0' [leading]; can include zero elsewhere)

with repetition \(\rightarrow\)

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/5,1/6,2/6} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 5\times6\times6 = 180 \text{ ways} \]

without repetition \(\rightarrow\)

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/5,1/5,2/4} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 5\times5\times4 = 100 \text{ ways.} \]

Note :- If not asked "with repetition / without repetition" & both answers are ready, then select the answer in which repetition is allowed.

(10) How many 6 digit numbers can be formed by the digits \(2,3,4,5,6,8\) when

(i) All the no's are divisible by 2

with repetition

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/6,1/6,2/6,3/6,4/6,5/4} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 6^5\cdot(4) \qquad \text{(4 = no of even digits)} \]

without repetition \(\rightarrow\)

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/1,1/2,2/3,3/4,4/5,5/4} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 4\times5! = 480 \text{ ways} \]

(ii) Every no. is greater than 400000

with repetition

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/4,1/6,2/6,3/6,4/6,5/6} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 4\times6^5 \qquad \text{(no of digits} \geq 4 \text{)} \]

without repetition

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/4,1/5,2/4,3/3,4/2,5/1} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = 4\times5! = 480 \text{ ways.} \]

(11) Ques: There are 10 people in a party. If each one shakes hand with the other exactly once, find the no. of handshakes?

\begin{tikzpicture}[scale=1.0]
  \foreach \x/\l [count=\i from 0] in {0/A,1/B,2/C,3/D,4/E,5/F,6/G,7/H,8/I,9/J} {
    \node at (\x,0) {$\l$};
  }
  \draw[->] (0,-0.15) .. controls (2,-1.0) .. (4,-0.15);
  \draw[->] (1,-0.15) .. controls (2.7,-0.9) .. (5,-0.15);
  \draw[->] (0,-0.15) .. controls (3,-1.3) .. (6,-0.15);
  \draw[->] (6.3,-0.15) .. controls (7,-0.7) .. (7.7,-0.15);
  \draw[->] (7,-0.15) .. controls (7.8,-0.6) .. (8.7,-0.15);
  \draw[->] (8,-0.15) .. controls (8.5,-0.45) .. (9,-0.15);
\end{tikzpicture}

\(A\) will shake hands with \(9\) people

\(B\) \(\longrightarrow\) \(8\) people

\(C\) \(\longrightarrow\) \(7\)

\(D\) \(\longrightarrow\) \(6\)

(and so on down to \(1\))

\[ \text{Total} = 9+8+7+6+5+4+3+2+1 = \frac{9\times10}{2} = 45 \text{ handshakes.} \]

or :- No of ways of selecting 2 people for handshake among 10.

\[ {}^{n}C_{2} = {}^{10}C_{2} = \]

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/10,1/9} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = \frac{9\times10}{2!} = 45 \]

(No of ways of arranging 2 people)

(12) If total no. of handshaking is 66, find the no. of people?

(a) 10    (b) 11    (c) 12    (d) 13

\[ {}^{n}C_{2} = 66 \;\Rightarrow\; \frac{n!}{(n-2)!\,2!} = 66 \] \[ \frac{n(n-1)}{2} = 66 \] \[ n(n-1) = 132 \] \[ n^2-n-132=0 \] \[ n = \frac{1\pm\sqrt{1+528}}{2} = \frac{1\pm\sqrt{529}}{2} = \frac{1+23}{2} \] \[ (n=12) \]

(13) If there are 10 points, find the no of diagonals?

\begin{tikzpicture}[scale=0.9]
  \draw (0,0) rectangle (1.2,1.2);
  \draw (0,0) -- (1.2,1.2);
  \draw (0,1.2) -- (1.2,0);
  \node[below] at (0.6,-0.3) {\small for $(n=4)$};
\end{tikzpicture}

\begin{tikzpicture}[scale=1.0]
  \node[circle,fill,inner sep=1.2pt,label=above:{\small 1}] (p1) at (0.4,1.4) {};
  \node[circle,fill,inner sep=1.2pt,label=right:{\small 2}] (p2) at (1.4,0.9) {};
  \node[circle,fill,inner sep=1.2pt,label=right:{\small 3}] (p3) at (1.2,-0.3) {};
  \node[circle,fill,inner sep=1.2pt,label=left:{\small 4}] (p4) at (0.0,-0.3) {};
  \node[circle,fill,inner sep=1.2pt,label=left:{\small 5}] (p5) at (-0.3,0.9) {};
  \draw (p1)--(p2)--(p3)--(p4)--(p5)--(p1);
  \draw[dashed] (p1)--(p3);
  \draw[dashed] (p1)--(p4);
  \draw[dashed] (p2)--(p4);
  \draw[dashed] (p2)--(p5);
  \draw[dashed] (p3)--(p5);
\end{tikzpicture}

diagonals \(=5\) : \((1\)-\(3,\,1\)-\(4)\), \((2\)-\(4,\,2\)-\(5)\), \((3\)-\(5)\)

/[A photo — a hand pointing at a whiteboard worked example — was pasted into the notebook at this point. It re-derives the \(n=4\) case: \({}^4C_2 = \dfrac{4\times3}{2}=6\), diagonals \(=6-4=2\), boxed alongside \({}^{n}C_2 - n = \dfrac{n(n-3)}{2}\).]/

Points \(1\)–\(10\): listing all pairs and striking out sides (adjacent pairs) and repeats —

\((1\)-\(2)^{\times},\ 1\)-\(3,\ 1\)-\(4,\ 1\)-\(5,\ 1\)-\(6,\ 1\)-\(7,\ 1\)-\(8,\ 1\)-\(9,\ (1\)-\(10)^{\times}\)

\((2\)-\(3)^{\times},\ 2\)-\(4,\ 2\)-\(5,\ 2\)-\(6,\ 2\)-\(7,\ 2\)-\(8,\ 2\)-\(9,\ 2\)-\(10,\ (2\)-\(1)^{\times}\)

\((3\)-\(4)^{\times},\ 3\)-\(5,\ 3\)-\(6,\ 3\)-\(7,\ 3\)-\(8,\ 3\)-\(9,\ 3\)-\(10,\ (3\)-\(1)^{\times},\ (3\)-\(2)^{\times}\)

...(and so on for points \(4\) through \(10\), each time crossing out the two adjacent sides and any pair already listed)

Using combinations, we need to choose two points from 10 points.

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/10,1/7} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ = \frac{70}{2!} \qquad \Leftarrow \text{No of arrangements of 2 obj.} \] \[ = 35 \]

/[As written: the sentence above the box announces "choose two points from 10" (\(={}^{10}C_2=45\)), but the box itself already applies the diagonal shortcut \(n(n-3)=10\times7=70\), giving \(70/2=35\) diagonals directly — not \({}^{10}C_2\). The final value (35) is correct for the decagon; the two lines just use different formulas.]/

For 'n' points we have

\[ \left.\frac{n\times(n-3)}{2}\right|_{n=4} = \frac{4\times1}{2} = 2 \; \checkmark \]

for \((n=5)\) : no of diagonals \(= \dfrac{5\times2}{2} = 5\;\checkmark\)

\[ \text{no. of diagonals} = \frac{n(n-3)}{2} \qquad \left(\text{Domain}: n-3\geq0,\ n\geq3\right) \]

for \(n=3\): diagonals \(=0\)

OR

\[ {}^{n}C_{2} - n = \frac{n!}{(n-2)!\,2!} - n = \frac{n(n-1)}{2} - n = \frac{n^2-n-2n}{2} = \frac{n^2-3n}{2} = \frac{n(n-3)}{2} \]

Ques: If there are 10 points, find the no. of diagonals?

Sol\(^n\):- 1 line \(\rightarrow {}^{n}C_2\)    \(\triangle \rightarrow {}^{n}C_3\)    For 'n' points :-

\[ \text{No of straight lines} = {}^{n}C_{2} \] \[ \text{No of Triangles} = {}^{n}C_{3} \] \[ \text{No of quadrilateral} = {}^{n}C_{4} \] \[ {}^{n}C_{2} = \text{Total no. of straight lines (including diagonal)} \] \[ \text{Normal straight lines} = n \qquad (\text{So No of diagonals} = {}^{n}C_2 - n) \]

(14) Out of 17 players, there are 5 bowlers. Find no of ways of selecting a team of 11 players which has 2 bowlers.

No of ways of selecting 2 bowlers from 5 \(= {}^{5}C_{2}\)

No of ways of selecting 9 batsman out of 12 \(= {}^{12}C_{9}\)

\[ \text{Total No of ways} = {}^{5}C_2\times{}^{12}C_9 = \frac{5!}{3!\,2!}\times\frac{12!}{9!\,3!} \] \[ = \frac{5\times4}{2}\times\frac{12\times11\times10}{3\times2} \] \[ = 20\times110 = \underline{2200} \]

(15) Ques: A committee of 3 members is to be formed out of 5 males and 4 females. Find how many committees can be formed consisting of -

(i) 2 Male and 1 female in the committee

(ii) No female

(iii) At least 1 female

(iv) At most 1 female

(i) No of ways of selecting 2 male out of 5 males \(= {}^{5}C_2\)

No of ways of selecting 1 female out of 4 \(= {}^{4}C_1\)

So Total no of ways of selecting committee

\[ = {}^{5}C_2\times{}^{4}C_1 = \frac{5!}{2!\,3!}\times\frac{4!}{1!\,3!} \] \[ = 4\times\frac{5\times4}{2} = 10\times4 = \underline{40} \]

(ii) No of ways of selecting 3 males out of 5 \(= {}^{5}C_3 = 5\times4/2 = 10\)

\(\quad\) \(\quad\) \(\quad\) \(\quad\) 0 females \(\quad\) \(\quad\) 4 \(= {}^{4}C_0 = 1\)

Total no. of ways \(= \underline{10}\)

(iii) \((F,M)\)   \(F+M=3\)

\((1,2)+(2,1)+(3,0)\)

\[ \left({}^{n}C_2 = \frac{n(n-1)}{2} = \frac{n!}{(n-2)!\,2!} = \frac{n(n-1)}{2}\right) \] \[ {}^{4}C_1\times{}^{5}C_2 + {}^{4}C_2\times{}^{5}C_1 + {}^{4}C_3\times{}^{5}C_0 = 4\times\frac{5\times4}{2} + \frac{4\times3}{2}\times5 + 4\times1 \] \[ = 40+30+4 = \underline{74} \]

(OR) No of ways of at least one Female

\[ = \text{Total} - \text{None female} \] \[ = {}^{N_m+N_F}C_3 - {}^{N_m}C_3 \] \[ = {}^{5+4}C_3 - {}^{5}C_3 = {}^{9}C_3 - {}^{5}C_3 = \frac{9\times8\times7}{3\times2} - \frac{5\times4}{2} = 84-10=\underline{74} \]

(iv) \(N(\text{At most 1 female}) = ?\)   \(5M,\ 4F\)   \((F+M=3)\)

\[ (M,F) = (3,0)+(2,1) \] \[ {}^{5}C_3\times{}^{4}C_0 + {}^{5}C_2\times{}^{4}C_1 = \frac{5\times4}{2}\times1 + \frac{5\times4}{2}\times4 \] \[ = 10(1+4) = \underline{50} \]

(16) Ques: How many different vegetables can be made from Cauliflower, tomatoes, onions, potatoes and carrots?

(a) 16   (b) 28   (c) 31 \(\checkmark\)   (d) 32

\[ N = {}^{5}C_1+{}^{5}C_2+{}^{5}C_3+{}^{5}C_4+{}^{5}C_5 \] \[ = 5 + \frac{5\times4}{2} + \frac{5\times4\times3}{3\times2} + 5 + 1 \] \[ = 5+10+10+5+1 \] \[ = 30+1 = \underline{31} \]

2018-June-Physics:

(17) Ques: In a group of 11 persons, each shakes hand with every other once and only once. What is the total number of such handshakes?

(a) 110   (b) 121   (c) 55 \(\checkmark\)   (d) 66

\[ \text{No of hand shakes} = \text{No of ways of selecting 2 people} \] \[ = {}^{11}C_2 = \frac{11\times10}{2} = \underline{55} \]

Dec-2017 (Physics):

(18) Ques: The number of 3 english letter words, having at least one consonant but not having two consecutive consonants is:

(a) 2205   (b) 3780 \(\checkmark\)   (c) 2730   (d) 3360

No of letters \(= 26\)

No of consonants \(= 26-5 = 21\)

\begin{tikzpicture}[scale=1.0]
  \foreach \i in {0,1,2} {
    \draw (\i,0) rectangle (\i+1,0.9);
  }
\end{tikzpicture}

\[ N = \text{Total ways} - N(\text{none consonant}) - N(2 \text{ consecutive consonant}) \] \[ = 26\times26\times26 - 5\times5\times5 - \ldots \]

/[This route is crossed out ("OR") in favour of the case-based method below.]/   (\(C\): consonant)

\((m)\)   \(V\;V\;C\)

\((n)\)   \(V\;C\;V\)

\(\;\;\;\;\;\,C\;V\;V\)      \(\Big\} \to N_1\)

\((m)\)   \(2C\): \(C\;V\;C\)      \(\to N_2\)

\((or)\)   \(3C\): \(C\;C\;C\) — 2 consecutive consonants, so this case won't contribute \(\to N_3 = 0\)

(Required) \(N_T = N_1+N_2+\cancel{N_3} = \;?\)   (Total ways)

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/5,1/5,2/21} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/5,1/21,2/5} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/21,1/5,2/5} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
\end{tikzpicture}

\[ N_1 = 5\times5\times21\times3 = 75\times21 = 25\times(20+1) = 1500+75 = 1575 \]

\begin{tikzpicture}[scale=1.0]
  \foreach \i/\txt in {0/21,1/5,2/21} {
    \draw (\i,0) rectangle (\i+1,0.9);
    \node at (\i+0.5,0.45) {$\txt$};
  }
  \node[above] at (0.5,1.0) {\small $C$};
  \node[above] at (1.5,1.0) {\small $V$};
  \node[above] at (2.5,1.0) {\small $C$};
\end{tikzpicture}

\[ N_2 = 21\times21\times5 = 441\times5 = 2205 \]

Required Ways \(= N_1+N_2 = 1575+2205 = \underline{3780}\)

Dec-2015 (Chemistry):

(19) Ques: The number of diagonals of a convex decodecagon (12-gon) is:

(a) 66   (b) 54 \(\checkmark\)   (c) 55   (d) 60

No of points: \(n=12\)

\[ \text{No of diagonals} = \text{Total lines} - \text{normal lines} \] \[ = {}^{n}C_2 - n \] \[ = \frac{n(n-3)}{2} = \frac{12\times9}{2} = \underline{54} \]

JUNE-2013:

(20) Ques: During a summer vacation, of 20 friends from a hostel, each wrote a letter to each of all others. The total number of letters written was -

(a) 20   (b) 400   (c) 200   (d) 380 \(\checkmark\)

Sol\(^n\):-

Any one will write 19 letters

So Total letters \(= 19\times20\)

\[ = \underline{380} \]

/[Unlike the handshake problems, writing a letter isn't mutual, so this uses \(19\times20\) directly rather than \({}^{20}C_2\).]/

(21) Qs: What is the number of distinct arrangements of the letters of the word "UGCCSIR" so that U and I can not together?

(a) 2520   (b) 720   (c) 1520   (d) 1800 \(\checkmark\)

\[ N(\text{'UI' can't come together}) = N_T - N(\text{UI come together}). \]

\(N_T = \) total no of words \(= \) "U G C C S I R" (positions 1 2 3 4 5 6 7)

\[ = \frac{7!}{2!} = 7\times6\times5\times4\times3 = 7\times360 = 2520 \]

No of ways in which UI come together

\[ = (UI)\;GCCSR \quad (\text{positions } 1\text{--}2\text{--}3\text{--}4\text{--}5\text{--}6) \] \[ = \frac{6!}{2!}\times2! = 6! = 6\times120 = 720 \] \[ N_{req} = N_T - 6! = 2520-720 = \underline{1800} \]

Ques: How many numbers between 99 and 1000 can be formed with the digits \(0,\!\!\not{\phantom{0}}\ 1,\!\!\not{\phantom{1}}\ 2,3,4\ \&\ 5\)

/[As written: '0,' and '1,' are struck through with an X in the notebook, but the working below still draws from a 6-digit pool — matching Q9's digit set \(\{0,1,2,3,4,5\}\) exactly, not the reduced \(\{2,3,4,5\}\) the strike-through would suggest.]/

Sol\(^n\):- \(99-1000\)

(without repetition)    rep. allowed

\[ \frac{5\times5\times4}{} = 100\ Ans. \qquad\qquad \frac{5\times6\times6}{} = 180\ Ans. \]