- mathmath333

There are 40 students .In how many ways can they be arranged to
form a team with 1 captain and 4 vice-captains ?

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- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{There are 40 students .In how many ways can they be arranged to}\hspace{.33em}\\~\\
& \normalsize \text{ form a team with 1 captain and 4 vice-captains ? }\hspace{.33em}\\~\\
\end{align}}\)

- steve816

40C1 * 40C4
I think that is the answer... btw, why are you only posting counting questions?

- mathmath333

why not 40C1 *40C4 *5!

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## More answers

- mathmath333

- steve816

Where did you get the 5! ?

- mathmath333

to permutate 5 students

- steve816

Actually, I think you might be right! 40C1 *40C4 *5!

- mathmath333

r u 100% sure

- steve816

Not a 100%, I struggled a little bit during probability unit too, but it seems right.

- BAdhi

when a captain is chosen, there will be 39 left so shouldnt it be
40C1*39C4

- mathmath333

but the 5 choosen students should be arranged in themselves

- ganeshie8

Alternative :
You can choose 5 ppl for the team in \(\dbinom{40}{5}\) ways
for each of that 5 member team, a captain can be chosen in \(\dbinom{5}{1}\) ways
so total possible teams = \(\dbinom{40}{5}*\dbinom{5}{1}\)

- ganeshie8

btw thats same as 40C1*39C4

- BAdhi

wow there are totally different ways to think of it.

- BAdhi

Since the arrangement inside the group is not relavant 5! is not required

- mathmath333

but in question the word 'arranged' is given

- ganeshie8

question is ambiguous
you can't do anything if the question is ambiguous

- ganeshie8

do you have answer ?

- mathmath333

yes 40 *39C4 is given answer

- ganeshie8

then read the question like this :
\[
\large \color{black}{\begin{align}
& \normalsize \text{There are 40 students. How many ways can a 5 member team }\hspace{.33em}\\~\\
& \normalsize \text{ with 1 captain and 4 normal players be formed ?}\hspace{.33em}\\~\\
\end{align}}
\]

- BAdhi

but in common sense, if we take a team, the team will not change if the team members are arranged in a different order. The team will be the same

- mathmath333

ok but still max people gave 40C1*39C4 at first glance

- ganeshie8

40C1*39C4
is same as
40*39C4

- mathmath333

i m getting suspeciuos of my different approach of understanding

- mathmath333

yea i know its same

- mathmath333

thnks

- Jhannybean

These questions are pretty fun to think about and solve.

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