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$\Delta =p!\left(p+1\right)!\left(p+2\right)!{\Delta }_1=2p!\left(p+1\right);\left(p+2\right)!=2{\left(p!\right)}^3.{\left(p+1\right)}^2.{\left(p+2\right)}^1\to \alpha +\beta =3+1=4$

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill p+1\hfill & \hfill \left(p+2\right)\left(p+1\right)\hfill \\ \hfill 1\hfill & \hfill p+2\hfill & \hfill \left(p+3\right)\left(p+2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+2\right)\left(-2\right)\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+3\right)\left(-2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|$

= 2(p + 3) – 2 (p + 2) = 2

$U=n{c}_{v}T$

$\begin{array}{l}=2\times \frac{3}{2}\times R\times 300\\ =7479J\end{array}$      

$B_A=\frac{{\mu }_{0}I\theta }{4\pi R}$

$\Rightarrow \mathrm{}\frac{B_A}{B_B}=\frac{I_A{\theta }_A{R}_B}{I_B{\theta }_B{R}_A}$

$\Rightarrow \frac{2\left(\frac{3\pi }{2}\right)\left(4\right)}{3\left. [\frac{5\pi }{3}\right]\left[2\right]}$

$\Rightarrow \mathrm{}\frac{6}{5}$ .

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 $A=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right],a_2,b_2,c_2\in \left\{0,1\right\}$

S = $a_1+a_2+a_3+b_1+b_2+b_3+c_1+c_2+c_3$ is prime

$0\le s\le 9$

Prime value = 2,3, 5, 7

$ForS=2,1+1+0+0+0+0+0+0+0\to \frac{9!}{2!7!}=36$

Total number of matrices

= 36 + 84 + 126 + 36 = 282

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When comparing BSc programmes such as those offered by IIHM Kolkata, students should consider key factors like course specialisation, industry exposure, faculty expertise, internship opportunities, placement success, and overall return on investment. The BSc in International Hospitality Management at IIHM Kolkata stands out for its strong industry links, global exposure, and affiliation with MAKAUT for degree validation.

Using newton’s law:

2mg – T = 4ma  - (1)

T – mg = ma       - (2)

From (1) & (2) :

$mg=5ma\Rightarrow a=\frac{g}{5}=2m/s^2$                

$T=mg+\frac{mg}{5}=\frac{6mg}{5}$                     

Sol. $\Rightarrow \mathrm{}{\stackrel{?}{\mathrm{V}}}_1=2\stackrel{?}{{\mathrm{U}}_{\mathrm{c}\mathrm{m}}}-\stackrel{?}{{\mathrm{U}}_1}\Rightarrow \mathrm{}2\left[\frac{\mathrm{m}/2{\mathrm{V}}_{\mathrm{o}}}{\frac{\mathrm{m}}{2}+\frac{\mathrm{m}}{3}}\right]-{\mathrm{V}}_{\mathrm{o}}$

$\Rightarrow \mathrm{}\frac{6}{5}{\mathrm{V}}_{\mathrm{o}}-{\mathrm{V}}_{\mathrm{o}}\Rightarrow \frac{{\mathrm{V}}_0}{5}$

$\Rightarrow \mathrm{}{\lambda }_{\mathrm{o}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{m}}{2}{\mathrm{V}}_{\mathrm{o}}\right)}\mathrm{}{\lambda }_{\mathrm{f}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{M}}{2}\frac{{\mathrm{V}}_{\mathrm{o}}}{5}\right)}$

$\Rightarrow \mathrm{\Delta }\lambda =\frac{8\mathrm{h}\mathrm{c}}{{\mathrm{m}\mathrm{V}}_{\mathrm{o}}}$

 $T_5{=}^n{C}_4{\left(2^{\frac{1}{4}}\right)}^{n-4}.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^4{=}^n{C}_4.2^{\frac{n-4}{4}}.\frac{1}{3}$

$T_6{=}^9{C}_5{\left(2^{\frac{1}{2}}\right)}^4.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^5{=}^9{C}_5.2.\frac{1}{3^{5/4}}=\frac{2}{3}.\frac{1}{3^{\frac{1}{4}}}.\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}$

Using conservation of momentum:

60 × v = (60 + 120) × 2

⇒  v = 6m/s

 ${\displaystyle \sum _{r=1}^{\infty }\frac{a+\left(r-1\right)d}{2^r}=4,4\left(a+d\right)=?}$

$\Rightarrow a{\displaystyle \sum _{r=1}^{\infty }{\left(\frac{1}{2}\right)}^r+d{\displaystyle \sum _{r=1}^{\infty }\left(r-1\right){\left(\frac{1}{2}\right)}^r=4}}$

$={\left(\frac{1}{2}\right)}^2{\displaystyle \sum _{k=0}^{\infty }k{\left(\frac{1}{2}\right)}^{k-1}}$

$=\frac{1}{4}.\frac{1}{{\left(1-\frac{1}{2}\right)}^2}$

= 1

a + d = 4

4 (a + d) = 16

(a, 4a, 7). (3, 1, 2b)= 0

3a + 4a – 14b = 0 a – 2b = 0 ……. (i)

(a, 4a, 7) . (b, a, 2) = 0

ab – 4a² + 14 = 0……… (ii)

(i) & (ii) =>2b² – 16b²+ 14 = 0

$\Rightarrow b^2=1\Rightarrow a=2b=\pm 2$

Plane : x – y + z = 0

P (4, 5, 1)

4 + 5 + 1 = 10

Yes, IIHM Kolkata offers a BSc programme in International Hospitality Management, including specialisations affiliated with MAKAUT such as BSc (H) in International Hospitality Management and Hospitality Administration. The course is of three years' duration, and admission is entrance-based, requiring candidates to clear the IIHM eCHAT exam followed by a Personal Interview (PI) round.

According to the 7th Central Pay Commission, the SSC JE salary under Pay Level 6 falls within a pay scale of INR 35,400 to INR 112,400. After deductions, in-hand salary ranges from INR 42,000 to 50,000.

Admission to the BA programme at IIHM Kolkata is strictly entrance-based. Candidates must appear for the IIHM eCHAT exam, followed by a personal interview. Only those with valid eCHAT scores and successful interview performance are shortlisted for final selection.

$K=\frac{\left(\frac{Q}{r}\right)\mathrm{\Delta }x}{\mathrm{\Delta }AT}$

$\begin{array}{ll}\Rightarrow & \frac{\left({\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}\right)\left(\mathrm{L}\right)}{\left({\mathrm{L}}^2\right)\left(\theta \right)\left(\mathrm{T}\right)}\\ \Rightarrow \mathrm{}& {\mathrm{M}}^1{\mathrm{L}}^{1-}{\mathrm{T}}^{-3}{\theta }^{-1}\end{array}$

 $\overline{x}=15\Rightarrow {\displaystyle \sum _{i=1}^{20}{x}_i=300}$

$\frac{{\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2}}{20}=9\Rightarrow {\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2=180}$

${\displaystyle \sum _{i=1}^{20}{\left(x_i+\alpha \right)}^2=178\times 20=3560}$

$4680+2\alpha \left(300\right)+20{\alpha }^2=3560$

${\alpha }^2+30\alpha +234-178=0$

= 2, 28

${\alpha }_{max}2={\left(-2\right)}^2=4$