Let X_n = {1, 2, 3, ldots, n} and let a subse | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $S_n$ be the number of subsets $A \subseteq \{1, 2, \ldots, n\}$ such that any two elements differ by at least $3$. Let $a_n$ be the number of

Vedclass · Accepted Answer

$p > q$ and $p - q = \frac{1}{10}$

Vedclass · Answer

(C) Let $S_n$ be the number of subsets $A \subseteq \{1, 2, \ldots, n\}$ such that any two elements differ by at least $3$. Let $a_n$ be the number of such subsets. For $n=10$,we calculate the total number of valid subsets $N$. Let $f(n)$ be the number of such subsets for $X_n$. The recurrence relation is $f(n) = f(n-1) + f(n-3) + 1$ (where $1$ accounts for the empty set). Calculating values: $f(0)=1, f(1)=2, f(2)=3, f(3)=4, f(4)=6, f(5)=9, f(6)=13, f(7)=19, f(8)=28, f(9)=41, f(10)=60$. Total subsets $N = 60$. Number of subsets containing $1$: If $1 \in A$,then $2, 3 
otin A$. We need to choose a subset from $\{4, 5, \ldots, 10\}$ such that elements differ by at least $3$. This is equivalent to choosing a subset from $\{1, 2, \ldots, 7\}$ with the same condition. Thus,$N(1 \in A) = f(7) = 19$. So,$p = \frac{19}{60}$. Number of subsets containing $2$: If $2 \in A$,then $1, 3, 4 
otin A$. We need to choose a subset from $\{5, 6, \ldots, 10\}$ such that elements differ by at least $3$. This is equivalent to choosing a subset from $\{1, 2, \ldots, 6\}$ with the same condition. Thus,$N(2 \in A) = f(6) = 13$. So,$q = \frac{13}{60}$. Therefore,$p > q$ and $p - q = \frac{19-13}{60} = \frac{6}{60} = \frac{1}{10}$.

column-$I$	column-$II$
$(A)$ $A$ line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $PQ=d$,then $d^2$ is	$(p)$ $-4$
$(B)$ The values of $x$ satisfying $\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}\left(\frac{3}{5}\right)$ are	$(q)$ $0$
$(C)$ Non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} \cdot \vec{b}=0$,$(\vec{b}-\vec{a}) \cdot(\vec{b}+\vec{c})=0$ and $2\|\vec{b}+\vec{c}\|=\|\vec{b}-\vec{a}\|$. If $\vec{a}=\mu \vec{b}+4 \vec{c}$,then the possible values of $\mu$ are	$(r)$ $4$
$(D)$ Let $f$ be the function on $[-\pi, \pi]$ given by $f(0)=9$ and $f(x)=\frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$ for $x \neq 0$. The value of $\frac{2}{\pi} \int_{-\pi}^\pi f(x) dx$ is	$(s)$ $5$
	$(t)$ $6$

Similar Questions

Two persons $A$ and $B$ take turns in throwing a pair of dice. The first person to throw a sum of $9$ with both dice will win the game. If $A$ throws first,then the probability that $B$ wins the game is:

If three boxes contain $3$ white and $1$ black,$2$ white and $2$ black,and $1$ white and $3$ black balls respectively,and one ball is chosen at random from each box,what is the probability of selecting $2$ white and $1$ black ball?

$A$ bag contains $2n$ coins,out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that a head appears in the toss is $\frac{41}{56}$,then the number of unfair coins in the bag is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module