3 Marks Question

Question 13 Marks

Using the letters of the word 'ARRANGEMENT' how many different words (using all letters at a time) can be made such that both $A$, both $E$, both $R$ and both $N$ occur together.

Answer

There are 11 letters in the word 'ARRANGEMENT' out of which $2 A^{\prime} s , 2 E ^{\prime}$ s and $2 N^{\prime}$ s and $2 R ^{\prime} s$
Considering both A , both E , both R and both N together, 8 letters should be counted as 4 .
So, there are total 7 letters (AA EE RR NN G N T)
These 7 letters can be arranged in 7 ! ways
$\begin{aligned} \text {Hence, total ways } & =7! \\ & =7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ & =5040 .\end{aligned}$

View full question & answer→

Question 23 Marks

A polygon has 35 diagonals. Find the number of its sides.

Answer

Let $n$ be the number of sides of a polygon and $D$ be the number of diagonals of that polygon.
We know that,
$
\begin{array}{l}
D={ }^n C_2-n= 35=\frac{n^2-3 n}{2}
\end{array}
$
$
\therefore \quad 35=\frac{n^2-3 n}{2}
$
$\begin{aligned} \Rightarrow & & n^2-3 n-70 & =0 \\ \Rightarrow & & (n-10)(n+7) & =0 \\ \Rightarrow & & n & =10,-7\end{aligned}$
Since, sides cannot be negative, therefore $n=10$. Hence, polygon is a decagon.

View full question & answer→

Question 33 Marks

Using the digits $0,1,2,2,3$, how many numbers greater than 2000 can be made ?

Answer

Total number of digits $=0,1,2,2,3$
Total number formed by these digits $=\frac{5!}{2!}=\frac{120}{2}$ $=60$
Total number formed by starting $0=\frac{4!}{2!}=\frac{24}{2}=12$
Total number formed by starting $1=\frac{4!}{2!}=\frac{24}{2}=12$
Total number formed greater than 2000
$=60-12-12=36$

View full question & answer→

Question 43 Marks

Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Answer

One ace will be selected from four aces and four cards will be selected from $(52-4)=48$ cards. If P is the required number of ways, then
$
\begin{aligned}
P & =C(4,1) \times C(48,4) \\
& =\frac{4!}{1!(4-1)!} \times \frac{48!}{4!(48-4)!} \\
& =\frac{4(3!)}{1!3!} \times \frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 1 \times 44!} \\
& =4 \times 2 \times 47 \times 46 \times 45 \\
& =778320 \text { ways. }
\end{aligned}
$

View full question & answer→

Question 53 Marks

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Answer

The number of ways of selecting 3 red balls out of 6 red balls $={ }^6 C_3$
The number of ways of selecting 3 white balls out of 5 white balls $={ }^5 C_3$
The number of ways of selecting 3 blue balls out of 5 blue balls $={ }^5 C_3$
The number of ways of selecting 3 red balls of each colour
$
\begin{array}{l}
={ }^6 C_3 \times{ }^5 C_3 \times{ }^5 C_3={ }^6 C_3 \times{ }^5 C_2 \times{ }^5 C_2 \\
=\frac{6 \times 5 \times 4}{1 \times 2 \times 3} \times \frac{5 \times 4}{1 \times 2} \times \frac{5 \times 4}{1 \times 2} \\
=20 \times 10 \times 10 \\
=2000
\end{array}
$

View full question & answer→

Question 63 Marks

Determine $n$, if ${ }^{2 n} C_3:{ }^n C_3=11: 1$.

Answer

Given, ${ }^{2 n} C_3:{ }^n C_2=12: 1$
$
\begin{array}{l}
\Rightarrow \quad \frac{2 n!}{3!(2 n-3)!} \div \frac{n!}{2!(n-2)!}=\frac{12}{1} \\
\Rightarrow \frac{2 n(2 n-1)(2 n-2)}{1 \times 2 \times 3} \div \frac{n(n-1)}{1 \times 2}=\frac{12}{1}
\end{array}
$
or,
$\begin{aligned} \frac{2 n(2 n-1) 2(n-1)}{6} & \times \frac{2}{n(n-1)}=\frac{12}{1} \\ & {\left[\because{ }^n C_r=\frac{n(n-1) \ldots(n-r+1)}{1 \times 2 \times 3 \ldots n}\right] }\end{aligned}$
or, $\quad \frac{4}{3}(2 n-1)=12$
$\Rightarrow \quad 2 n-1=9$
$\Rightarrow \quad 2 n=10$
or, $n=5 .$

View full question & answer→

Question 73 Marks

If ${ }^n P_4:{ }^n P_2=12$, find $n$.

Answer

Given, $\frac{{ }^" P_4}{{ }^" P_2}=12$
$\begin{array}{lc}\Rightarrow & { }^n P_4=12^n P_2 \\ \Rightarrow & \frac{n!}{n-4!}=12 \frac{n!}{n-2!} \\ \Rightarrow & \frac{n-2!}{n-4!}=12\end{array}$
$\Rightarrow \frac{(n-2)(n-3)(n-4)!}{n-4!}=12$
$\Rightarrow \quad n^2-5 n+6=12$
$\Rightarrow \quad n^2-5 n-6=0$
$\Rightarrow \quad n^2-6 n+n-6=0$
$\Rightarrow \quad n(n-6)+1(n-6)=0$
$\Rightarrow \quad(n-6)(n+1)=0$
$\Rightarrow \quad n=-1$ or $n=6$
$\because n$ cannot be negative
$\therefore n=6$

View full question & answer→

Applied Maths 3 Marks Question

Test yourself on this topic