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MATHS 1 Marks Question

PART - 1 CH - 6 Permutations and Combinations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 16 questions.

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Question 11 Mark
If ${ }^n C_{10}={ }^n C_{12}$ then find the value of $n$ and then write the value of ${ }^n C _5$.
Answer
${ }^n C_{10}={ }^n C_{12}$
$10=12$ is not possible.
Then $n=10+12=22$
$\begin{aligned}\therefore{ }^{22} C_5 & =\frac{22 \times 21 \times 20 \times 19 \times 18}{1 \times 2 \times 3 \times 4 \times 5} \\& =26334\end{aligned}$
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Question 21 Mark
There are 4 coats, 5 shirts and 6 hats among three people. How many ways can they wear them?
Answer
Required ways
$
\begin{array}{l}
={ }^4 P_3 \times{ }^5 P_3 \times{ }^6 P_3 \\
=4.3 .2 \times 5.4 .3 \times 6.5 .4 \\
=172800
\end{array}
$
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Question 31 Mark
18 students attended the student union meeting of a school. Each shakes hands with the other only once. Find the total number of handshakes.
Answer
Number of students $=18$
Here each student shakes hands with the other student only once. We have to get the number of two groups each out of 18 at a time.
Therefore, number of people shaking hands
$\begin{array}{l}={ }^{18} C_2=\frac{18 \times 17}{1 \times 2} \\=9 \times 17 \\=153\end{array}$
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Question 41 Mark
In how many ways can 6 apples be divided among 2 boys, there is no restriction on every boy getting an apple?
Answer
Each apple can be given to any ot the 3 boys and this can happen in 3 ways
Therefore total number of methods
$\begin{array}{l}=3^6 \\=3 \times 3 \times 3 \times 3 \times 3 \times 3 \\=27 \times 27 \\=729\end{array}$
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Question 51 Mark
Prove that :
${ }^n P_n=6 .{ }^n P_{n-3}$
Answer

$\begin{aligned} \text { LHS }= & { }^n P _n=\frac{n!}{(n-n)!}=\frac{n!}{0!}=n! \\ \text { RHS }= & 6 \cdot{ }^n P _{n-3}=6 \times \frac{n!}{(n-n+3)!}=\frac{6 n!}{3!} \\ & =\frac{6 n!}{6}=n!\end{aligned}$
$\quad\quad$LHS $=$ RHS$\quad$Hence proved.
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Question 61 Mark
In how many ways can 4 persons be selected out of 12 persons with the condition that a specific person must be included?
Answer

$\begin{aligned}{ }\text {Number of groups}\quad^{11} C_3 & =\frac{11 \times 10 \times 9}{1 \times 2 \times 3} \\& =165\end{aligned}$
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Question 71 Mark
A person wants to invite six friends. In how many ways can he sent invitation to them if he hires three servants?
Answer
Since a card can be sent by any one of the three servants, hence the number of ways to send invitation letter to the first friend $=3$
Similarly, there are 3 ways to send invitation letter to six friends.
Hence, the number of required methods
$\begin{array}{l}=3 \times 3 \times 3 \times 3 \times 3 \times 3 \\=3^6=729 .\end{array}$
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Question 81 Mark
If $(n+2)!=20 n!$ then find the value of $n$.
Answer

$\begin{aligned} & & (n+2)! & =20 n! \\ \Rightarrow & & (n+2)(n+1) n! & =20 n! \\ \Rightarrow & & (n+2)(n+1) & =20\end{aligned}$
$\begin{array}{l}\Rightarrow \quad n^2+3 n-18=0 \\\Rightarrow \quad(n+6)(n-3)=0 \\\Rightarrow \quad n=3,-6 \\n=-6 \text { is not possible, so } n=3.\end{array}$
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Question 91 Mark
How many ways will there be for a student to select 5 subject out of 9 subject, if two (2) subjects are compulsory?
Answer
${ }^7 C _3=\frac{7.6 .5}{3.2 .1}=35$.
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Question 101 Mark
In how many ways can 5 boys and 5 girls be seated around a round table such that no two girls sit together?
Answer
$\lfloor\underline{5}\cdot\lfloor\underline{4}$
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Question 111 Mark
In how many ways can 10 people be seated around a round table provided that no two neighbours have the same neighbours?
Answer
$\lfloor\underline{9}=362880$.
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Question 121 Mark
In how many ways can a team of 9 out of 12 cricket players be formed while the captain always remain?
Answer
$\begin{array}{l}{ }^{11} C _8=\frac{11 \times 10 \times 9}{3.2 .1} \\ =11 \times 5 \times 3=165\end{array} \quad\left[\because{ }^{11} C _8={ }^{11} C _3\right]$
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Question 151 Mark
If out of $n$ objects, $p$ objects are of one type, 4 objects are of second type, $r$ objects are of third type and other are different, then write the number of permutations.
Answer
$\frac{n!}{p!q!r!}$
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Question 161 Mark
Write the LCM of $\lfloor\underline{4},\lfloor\underline{5}$ and $\lfloor\underline{6}$.
Answer
$LCM =\lfloor\underline{6}$
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