Jee Main Math Class 11 Permutations And Combinations JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: 414

Explanation: Case-I: $1 \rightarrow7$ times and $-1 \rightarrow2$ times number of possible marrix $= \; \frac{9 !}{7 ! 2 !}=36$ Case-II: $1 \rightarrow6$ times, $-1 \rightarrow1$ times and $0 \rightarrow2$ times number of possible marrix $= \; \frac{9 !}{6 ! 2 !}=252$ Case-III: $1 \rightarrow5$ times and $0 \rightarrow4$ times number of possible marrix $= \; \frac{9 !}{5 ! 4 !}=126$ Hence total number of all such matrix $A=414$

Correct Answer: (b)

Explanation: $x+y+z=15$ $\; \text{Total no. solution} \; = {}^{15+3-1}C_3 = 136 \ldots \; \text{(1)}$ $\; \text{Let} \; x=y \neq z$ $2x+z=15 \Rightarrow z=15-2t$ $\Rightarrow r \in \{0,1,2,\ldots 7\}-\{5\}$ $\therefore 7 \; \text{solutions}$ $\therefore \; \text{there are} \; 21 \; \text{solutions in which exactly}$ $\; \text{Two of} \; x, y, z \; \text{are equal \ldots (2)}$ $\; \text{There is one solution in which} \; x=y=z \ldots \; \text{(3)}$ $\; \text{Required answer} \; =136-21-1=144$

Correct Answer: (c)

Explanation: Words starting with A,L,M $=\frac{4!}{2!}+4!+\frac{4!}{2!}=48$ Words starting with SA,SL= $\frac{3!}{2!}+3!=9$ ⇒ Rank of the word SMALL = 58

Correct Answer: 11

Explanation: 4 dice are independently thrown. Each die has probability to show 3 or 5 is $p = \frac{2}{6} = \frac{1}{3}$ $\therefore q = 1 - \frac{1}{3} = \frac{2}{3}$ (not showing 3 or 5) Experiment is performed with 4 dices independently. ∴ Their binomial distribution is $(q + p)^{4} = q^{4} + {}^{4}C_{1} q^{3} p + {}^{4}C_{2} q^{2} p^{2}$ $+ {}^{4}C_{3} q p^{3} + {}^{4}C_{4} p^{4}$ $\therefore$ In one throw of each dice probability of showing 3 or 5 at least twice is $= p^{4} + {}^{4}C_{3} q p^{3} + {}^{4}C_{2} q^{2} p^{2}$ $= \frac{33}{81}$ $\therefore$ Such experiment performed 27 times $\therefore$ so expected out comes $= np$ $= \frac{33}{81} \times 27$ $= 11$

Correct Answer: (c)

Explanation: ${}^{n}P_{r}={}^{n}P_{rtl}\Rightarrow \frac{n!}{(n-r)!}=\frac{n!}{(n-r-1)!}$ $\Rightarrow (n-r)=1$ ........(1) ${}^{n}C_{r}={}^{n}C_{r-1}$ $\Rightarrow \frac{n!}{r!(n-r)!}=\frac{n!}{(r-1)!(n-r+1)!}$ $r\frac{1}{r(n-r)!}=\frac{1}{(n-r+1)(n-r)!}$ $\Rightarrow n-r+1=r$ $\Rightarrow n+1=2r$ .......(2) (1) $\Rightarrow 2r-1-r=1\Rightarrow r=2$