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mathmath333
 one year ago
Counting question
mathmath333
 one year ago
Counting question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{There are 7 people and 4 chairs. }\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can the chairs be occupied.}\hspace{.33em}\\~\\ \end{align}}\)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3We can do that in two steps. 1st step: Choosing 4 people out of 7(Combination ) 2nd step: Seating them.(Permutation)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3In my method the answer will be this.(Giving the same as nnesha's result): \(\huge ^7C_4\times 4!\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.11st step: Choosing 4 people out of 7(Combination ) 2nd step: Seating them.(Permutation) How to implement this

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3\[\huge \ldots=P_{(7,4)} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4i 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4the problem is equivalent to that of finding number of 4 letters words using 7 different letters

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i think it as only 4 persons can be seated and the 4 persons can be arranged in 4! ways

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3Simplicity @ganeshie8 :D lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.44! is true if you just have 4 chairs

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3....and only 4 people..

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yes i just have 4 chairs

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4** 4! is true if you just have 4 chairs and 4 people

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3You can take a simple example like 4 people and 3chairs and get the feeling @mathmath333

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3people ABCD ok? All possible ways: ABC ACB ABD ADB ACD ADC .... ... Fill in the blanks and you've get the feeling....

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is there any website that shows permutations and combinations list

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3python has got the function permutations in itertools module.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1ABC ACB ABD ADB ACD ADC BCD BDC

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you may use this for combinations https://jsfiddle.net/ganeshie8/r2hjr3js/4/

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1it 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}

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4right you should get 7C4 = 35 combinations

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4for permutations, notice that each of that combination above can be permuted in 4! ways so number of permutations = 35*4!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is this question same as "In how many ways can a 4 digit number be formed using digits "1,2,3,4,5,6,7"

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3yep... If you aren't allowed to use the same number twice.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1from 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you're not alone, everybody is facing that issue @Astrophysics is there any fix yet ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1lol how can so many people tolerate this

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1previuosly it happened only once in an hour but recently it get to on every long text i type
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