Permutations Help

[yogibear13]

We have four Attempts as a Group or a Set ; Each attempt has two outcomes ; either True or False but not both. How many combinations of these Groups or Sets( of 4) can be formed using either True or False for each attempt ? for example ONE of the combinations could be :-
Attempt 2 Attempt 3 Attempt 4 Attempt 5
T F F T

 

[Deleted member 4993]

We have four Attempts as a Group or a Set ; Each attempt has two outcomes ; either True or False but not both. How many combinations of these Groups or Sets( of 4) can be formed using either True or False for each attempt ? for example ONE of the combinations could be :-
Attempt 2 Attempt 3 Attempt 4 Attempt 5
T F F T

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[Steven G]

Can't you even try to list them? What have you tried other then to expect helpers on this forum to help. I guess I am in a good mood since I will list the outcomes for you.

TTTT---0 falses

TTTF, TTFT, TFTT, FTTT ---1 FALSE

TTFF, TFTF, TFFT, FTFF, FFTF, FTFF -2 FALSES

FFFT, FFTF, FTFF, FTFF--3 FALSES

FFFF--4 FALSES.

I know you are only in arithmetic, but you should be able to do these problems.

 

[yogibear13]

Many Thanks. I was afraid that it would have numerous combinations so wanted to know how many before I try to list them.

 

[pka]

We have four Attempts as a Group or a Set ; Each attempt has two outcomes ; either True or False but not both. How many combinations of these Groups or Sets( of 4) can be formed using either True or False for each attempt ? for example ONE of the combinations could be :-
Attempt 2 Attempt 3 Attempt 4 Attempt 5
T F F T

For any binary event such as: heads or tails; true or false; on or off; in or out; etc
If we have \(n\) attempts then the outcome table has \(2^n\) rows and \(n\) columns.
For example, say we flip a coin six times then there \(2^6=64\) rows and six columns.
To construct the table start with 32 H's followed by 23 T's for the first column.
Then the second column we have sixteen H's followed by sixteen T's followed by sixteen H's followed by sixteen T's.
Then repeat that pattern dividing the blocks of H's & T's in in preceding preceding column by two.
The last column the H's & T's alternate. Here is the in out table for \(n=4\).
\(\begin{array}{*{20}{c}}
I&I&I&I \\
I&I&I&O \\
I&I&O&I \\
I&I&O&O \\
I&O&I&I \\
I&O&I&O \\
I&O&O&I \\
I&O&O&O \\
O&I&I&I \\
O&I&I&O \\
O&I&O&I \\
O&I&O&O \\
O&O&I&I \\
O&O&I&O \\
O&O&O&I \\
O&O&O&O
\end{array}\)