# ISI2014-DCG-41

1 vote
157 views

The number of permutations of the letters $a, b, c$ and $d$ such that $b$ does not follow $a,c$ does not follow $b$, and $c$ does not follow $d$, is

1. $11$
2. $12$
3. $13$
4. $14$

recategorized

$\mathbf{\underline{Answer: A}}$

$\underline{\mathbf{Explanation:}}$

Total number of permutations of $\color{darkorange}{\mathbf{a, b, c, d}} = 4!= 24$

Number of permutations of $\color{blue}{\mathbf{ab, c, d} }= 3!= 6$
Number of permutations of $\color{blue}{\mathbf{a, bc, d}} = 3!= 6$
Number of permutations of $\color{blue}{\mathbf{a ,b ,cd} }= 3!= 6$

Number of permutations of $\color{green}{\mathbf b} \;\text{and}\; \color{green}{\mathbf c} = 2!= 2$
Number of permutations of $\color{green}{\mathbf c} \;\text{and} \;\color{green}{\mathbf d} = 2!= 2$
Number of permutations of $\color{green}{\mathbf d} \;\text{and} \;\color{green}{\mathbf b}= 2! =2$

Number of permutations of $\color{green}{\mathbf{b,c}} \;\text{and} \; \color{green}{\mathbf d} = 1! = 1$

Hence, required permutations are:

$\mathbf{4! - 3! - 3! - 3! + 2! + 2! + 2! - 1! = \underline{11}}$

$\therefore$ Option $\mathbf A$ is the correct answer.

edited by
0
Have you  tried to solve it using inclusion-exclusion principle? Please tell me how do you solve such questions, I am always stuck.
0
Just take a simple test case and analyse it in and out, this is the best way to answer such questions easily.
1
shouldnt it be dc instead of cd?

## Related questions

1
346 views
Let $(1+x)^n = C_0+C_1x+C_2x^2+ \dots + C_nx^n$, $n$ being a positive integer. The value of $\left( 1+\dfrac{C_0}{C_1} \right) \left( 1+\dfrac{C_1}{C_2} \right) \cdots \left( 1+\dfrac{C_{n-1}}{C_n} \right)$ is $\left( \frac{n+1}{n+2} \right) ^n$ $\frac{n^n}{n!}$ $\left( \frac{n}{n+1} \right) ^n$ $\frac{(n+1)^n}{n!}$
$^nC_0+2^nC_1+3^nC_2+\cdots+(n+1)^nC_n$ equals $2^n+n2^{n-1}$ $2^n-n2^{n-1}$ $2^n$ none of these
Consider $30$ multiple-choice questions, each with four options of which exactly one is correct. Then the number of ways one can get only the alternate questions correctly answered is $3^{15}$ $2^{31}$ $2 \times \begin{pmatrix} 30 \\ 15 \end{pmatrix}$ $2 \times 3^{15}$
The following sum of $n+1$ terms $2 + 3 \times \begin{pmatrix} n \\ 1 \end{pmatrix} + 5 \times \begin{pmatrix} n \\ 2 \end{pmatrix} + 9 \times \begin{pmatrix} n \\ 3 \end{pmatrix} + 17 \times \begin{pmatrix} n \\ 4 \end{pmatrix} + \cdots$ up to $n+1$ terms is equal to $3^{n+1}+2^{n+1}$ $3^n \times 2^n$ $3^n + 2^n$ $2 \times 3^n$