Two related Permutations/Combinations Questions

[kaebun]

The first one is a coin is tossed 20 times and the heads and tails sequence is recorded. From among the sequences of heads and tails, how many have exactically 7 heads? my answer in \(\displaystyle 2^20\) but I have a feeling that this is not completly right, I think there is another step but I'm not sure what it would be...

The second one is Six seniors at a school meet the qualifications for a scholarship for a university. the universiy allows the school to nominate up to three canidates and the school alway nominates at least one. How many different choices could the commite make?
I'm realy not sure what to do on this one,I think my main confusion is how to use all three numbers
Thank you for your time.

 

[pka]

The first one is a coin is tossed 20 times and the heads and tails sequence is recorded. From among the sequences of heads and tails, how many have exactly 7 heads?

How many ways can we arrange a string of 7H’s & 13T’s?
\(\displaystyle \L
\frac{{20!}}{{7! \cdot 13!}}\)

Six seniors at a school meet the qualifications for a scholarship for a university. The university allows the school to nominate up to three candidates and the school always nominates at least one. How many different choices could the committee make?

\(\displaystyle \L
\sum\limits_{k = 1}^3 {\left( \begin{array}{l}
6 \\
k \\
\end{array} \right)}\)

 

[soroban]

Hello, kaebun!

The first problem is badly worded . . .

A coin is tossed 20 times and the heads and tails sequence is recorded. ?
From among the sequences of heads and tails, how many have exactly 7 heads?

My answer is \(\displaystyle 2^{20}\) but I have a feeling that this is not completly right. . . . You're right: it's wrong.

If a coin is tossed 20 times, the result could be: HHTHTTHHTHTHTHHTHTTT
"From among the sequences"? . . . there's only one sequence!


\(\displaystyle 2^{20}\) is the number of different possible sequences for the 20 tosses.

Evidently, the problem is referring to a theoretical experiment.
\(\displaystyle \;\;\)"If a coin is tossed 20 times and we consider all 1,048,576 possible outcomes,
\(\displaystyle \;\;\)how many of them have exactly 7 heads?"


Obviously, your answer is wrong.
The number "7" has to enter into the calculation somewhere, right?

The answer is: \(\displaystyle \,\begin{pmatrix}20\\7\end{pmatrix}\:=\:\frac{20!}{7!\cdot13!}\:=\:77,520\)

 

[kaebun]

Okay i understand the first problem perfectly now
but I have no clue what this means...

\(\displaystyle \L
\sum\limits_{k = 1}^3 {\left( \begin{array}{l}
6 \\
k \\
\end{array} \right)}\)

 

[pka]

Well the school can choose 1, the school can choose 2, or the school can choose 3.
This is 6 choosing k: \(\displaystyle \left( \begin{array}{l}
6 \\
k \\
\end{array} \right) = \frac{{6!}}{{\left( {k!} \right)\left( {(6 - k)!} \right)}}\).

The add them up.

 

[kaebun]

thank you! that makes sence