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A coin is tossed $10$ times.

a) How many different sequences of heads and tails are possible?

b) How many different sequences containing $6$ heads and $4$ tails are possible?

c) What is the probability of getting $6$ heads and $4$ tails?

I genuinely just don't know what to do: I've tried $10!$ and $\binom{10}{2}$ but I'm clueless.

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    Answer for $2a)$ is $2^{10}=1024$ – 2017-01-12
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    @Peter why do you put it the power of 10? – 2017-01-12
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    If I have a die and I roll it some arbitrary amount, say 3 times. How many different sequences are possible? Can you solve that one? – 2017-01-12
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    if you had one toss, how many sequences are possible? Can you extend that to 2,3,4... tosses? – 2017-01-12
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    Each throw can be heads or tails, so the numbers of possibilities multiply. – 2017-01-12
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    @inandouto'mind oh it's 6^3 that makes sense now, thanks!! – 2017-01-12
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    I could be wrong but I think maybe you should try 2^x. – 2017-01-12
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    Answer for $2b)$ is $\binom{10}{4}=210$. Now , $2c)$ should be no problem anymore. – 2017-01-12
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    6 heads and 4 tails, how many ways can you arrange 4 withing 6 spaces, or 6 within 10 spaces? It's related to one of the things you mentioned you tried. – 2017-01-12
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    @Peter why is it 10C4 though? I get the 10 because there's 10 tosses, but why the 4? – 2017-01-12
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    @KrishaanKhubchand Count the ways to select 4 from 10 coins to be tails, and the others heads. – 2017-01-12
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    To choose $4$ out of $10$, we have $10\cdot9\cdot8\cdot7$ possibilities, but $4\cdot3\cdot2\cdot1$ combinations give the same sequence (For example, the choices $2,4,6,8$ and $2,8,6,4$ do not differ because it onlay matters which positions are chosen). If we fix the positions of tails, the positions of heads are determined as well. Hence , the number of possibilities is $\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}=\binom{10}{4}$, in general, there are $\binom{n}{k}$ possibilities to choose $k$ out of $n$ – 2017-01-12

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Let's put the ideas in the comments together (particularly, the ones by @Peter). I'll begin by solving a similar problem: finding the probability of getting exactly 3 Heads in five tosses of a fair coin.

(a) Possible sequences: $2^5 - 32.$ There are two possibilities on each of five tosses. Sample space $S$ can be represented as follows. There are $1 + 5 + 10 + 10 + 5 + 1 = 32$ elements in the sample space.

HHHHH
HHHHT HHHTH HHTHH HTHHH THHHH
HHHTT HHTHT HTHHT THHHT HHTTH HTTHH TTHHH HTHTH THTHH THHTH
TTTHH TTHTH THTTH HTTTH TTHHT THHTT HHTTT THTHT HTHTT HTTHT
TTTTH TTTHT TTHTT THTTT HTTTT
TTTTT

(b) Sequences with exactly three Heads: ${5 \choose 3} = 10.$ There are five 'slots' to fill, choose three of them for Heads in ${5 \choose 3}$ ways, put Tails in the remaining two slots. These are the ten sample elements on the third row above.

HHHTT HHTHT HTHHT THHHT HHTTH HTTHH TTHHH HTHTH THTHH THHTH

(c) Probability of getting exactly three Heads in five tosses: ${5 \choose 3}/2^5 = 10/32 =5/16.$

In a similar way, except that I don't have the patience to type the 1024 sample elements, the probability of getting exactly six heads in ten tosses is ${10 \choose 6}/2^{10}.$

Addendum: If you haven't already seen it, your course should soon cover the binomial distribution. The number $X$ of heads in $n = 10$ tosses of a fair ($p = 1/2$) coin has a binomial distribution. Specifically, $X \sim Binom(n = 10, p = 1/2).$

The formula for the PDF of a binomal distribution is $$P(X = k) = {n \choose k}p^k(1-p)^{n-k},$$ for $k = 0, 1, \dots, n.$ In case $p = 1/2 = 1 - p,$ this formula simplifies to $P(X = k) = {n \choose k}/2^n.$

Below is a bar graph of $Binom(10,1/2),$ in which the bar for $P(X = 6)$ is emphasized in a different color.

enter image description here