The conflict between generalists and specialists in public administration centers on the scope of expertise. Generalists advocate for a broad understanding of governance, promoting flexibility and adaptability, while specialists emphasize in-depth knowledge and technical skills in specific fields. The First Administrative Reforms Commission (ARC) recommended balancing both perspectives, advocating for the development of generalist officers with specialized training, enhancing inter-disciplinary cooperation, and fostering a collaborative environment. The ARC emphasized the need for specialists within administrative structures but urged that they operate under the leadership of generalists to ensure comprehensive governance and effective policy implementation.

Judicial control in India primarily manifests through constitutional review, writ jurisdiction, and public interest litigation (PIL). The Supreme Court and High Courts can invalidate laws and executive actions that violate the Constitution. Writs such as habeas corpus, mandamus, prohibition, and certiorari ensure protection of fundamental rights. PIL enhances access to justice, allowing individuals and NGOs to file cases for societal issues. However, challenges include judicial overreach, backlog of cases, and the need for judicial accountability. Despite these issues, judicial control remains vital for upholding the rule of law and protecting citizens’ rights in a democratic society.

The Graicunas formula for calculating the span of control is expressed as:

$S=N+N(N-1)\mathrm{/}2+N(N-1)(N-2)\mathrm{/}6S\; =\; N\; +\; N(N-1)/2\; +\; N(N-1)(N-2)/6$

S=N+N(N−1)/2+N(N−1)(N−2)/6

where

$SS$

S is the total number of relations, and

$NN$

N is the number of subordinates.

For

$N=5N\; =\; 5$

N=5:

1. Direct relations:
   $55$ 

   5

2. Group relations:
   $5(5-1)\mathrm{/}2=5\times 4\mathrm{/}2=105(5-1)/2\; =\; 5\; \backslash times\; 4/2\; =\; 10$ 

   5(5−1)/2=5×4/2=10

3. Functional relations:
   $5(5-1)(5-2)\mathrm{/}6=5\times 4\times 3\mathrm{/}6=105(5-1)(5-2)/6\; =\; 5\; \backslash times\; 4\; \backslash times\; 3/6\; =\; 10$ 

   5(5−1)(5−2)/6=5×4×3/6=10

Total relations:

$S=5+10+10=25S\; =\; 5\; +\; 10\; +\; 10\; =\; 25$

S=5+10+10=25

Thus, the total number of relations for 5 subordinates is 25.