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\(\large \color{black}{\begin{align} & \normalsize \text{There are 7 people and 4 chairs. }\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can the chairs be occupied.}\hspace{.33em}\\~\\ \end{align}}\)
We can do that in two steps. 1st step: Choosing 4 people out of 7(Combination ) 2nd step: Seating them.(Permutation)
is it 4!

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In my method the answer will be this.(Giving the same as nnesha's result): \(\huge ^7C_4\times 4!\)
1st step: Choosing 4 people out of 7(Combination ) 2nd step: Seating them.(Permutation) How to implement this
\[\huge \ldots=P_{(7,4)} \]
i like amilapsn's method, but here is an alternative that Nnesha is talking about : you look at 4 chairs, the first chair can have any one of the 7 persons : 7 ways after that, the second chair can have any one of the remaining 6 persons : 6 ways after that, the third chair can have any one of the remaining 5 persons : 5 ways after that, the fourth chair can have any one of the remaining 4 persons : 4 ways so total seating arrangements = 7*6*5*4
the problem is equivalent to that of finding number of 4 letters words using 7 different letters
i think it as only 4 persons can be seated and the 4 persons can be arranged in 4! ways
Simplicity @ganeshie8 :D lol
4! is true if you just have 4 chairs
....and only 4 people..
yes i just have 4 chairs
** 4! is true if you just have 4 chairs and 4 people
You can take a simple example like 4 people and 3chairs and get the feeling @mathmath333
people ABCD ok? All possible ways: ABC ACB ABD ADB ACD ADC .... ... Fill in the blanks and you've get the feeling....
is there any website that shows permutations and combinations list
python has got the function permutations in itertools module.
ABC ACB ABD ADB ACD ADC BCD BDC
you may use this for combinations https://jsfiddle.net/ganeshie8/r2hjr3js/4/
nice
it gave this for abcdefg and 4 comb {a,b,c,d} {a,b,c,e} {a,b,d,e} {a,c,d,e} {b,c,d,e} {a,b,c,f} {a,b,d,f} {a,c,d,f} {b,c,d,f} {a,b,e,f} {a,c,e,f} {b,c,e,f} {a,d,e,f} {b,d,e,f} {c,d,e,f} {a,b,c,g} {a,b,d,g} {a,c,d,g} {b,c,d,g} {a,b,e,g} {a,c,e,g} {b,c,e,g} {a,d,e,g} {b,d,e,g} {c,d,e,g} {a,b,f,g} {a,c,f,g} {b,c,f,g} {a,d,f,g} {b,d,f,g} {c,d,f,g} {a,e,f,g} {b,e,f,g} {c,e,f,g} {d,e,f,g}
right you should get 7C4 = 35 combinations
for permutations, notice that each of that combination above can be permuted in 4! ways so number of permutations = 35*4!
is this question same as "In how many ways can a 4 digit number be formed using digits "1,2,3,4,5,6,7"
yep... If you aren't allowed to use the same number twice.
ok thnx
from the past 2 weeks i m getting this very annoying bug I have to reload the page every time to type and my question gets changed to this page
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you're not alone, everybody is facing that issue @Astrophysics is there any fix yet ?
me too...
lol how can so many people tolerate this
previuosly it happened only once in an hour but recently it get to on every long text i type

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