Counting question

- mathmath333

Counting question

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- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{There are 7 people and 4 chairs. }\hspace{.33em}\\~\\
& \normalsize \text{In how many ways can the chairs be occupied.}\hspace{.33em}\\~\\
\end{align}}\)

- amilapsn

We can do that in two steps.
1st step: Choosing 4 people out of 7(Combination )
2nd step: Seating them.(Permutation)

- mathmath333

is it 4!

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## More answers

- amilapsn

In my method the answer will be this.(Giving the same as nnesha's result):
\(\huge ^7C_4\times 4!\)

- mathmath333

1st step: Choosing 4 people out of 7(Combination )
2nd step: Seating them.(Permutation)
How to implement this

- amilapsn

\[\huge \ldots=P_{(7,4)} \]

- ganeshie8

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

- ganeshie8

the problem is equivalent to that of
finding number of 4 letters words using 7 different letters

- mathmath333

i think it as only 4 persons can be seated and the 4 persons can be arranged in 4! ways

- amilapsn

Simplicity @ganeshie8 :D lol

- ganeshie8

4! is true if you just have 4 chairs

- amilapsn

....and only 4 people..

- mathmath333

yes i just have 4 chairs

- ganeshie8

** 4! is true if you just have 4 chairs and 4 people

- amilapsn

You can take a simple example like 4 people and 3chairs and get the feeling @mathmath333

- amilapsn

people ABCD ok?
All possible ways:
ABC
ACB
ABD
ADB
ACD
ADC
....
...
Fill in the blanks and you've get the feeling....

- mathmath333

is there any website that shows permutations and combinations list

- amilapsn

python has got the function permutations in itertools module.

- mathmath333

ABC
ACB
ABD
ADB
ACD
ADC
BCD
BDC

- ganeshie8

you may use this for combinations
https://jsfiddle.net/ganeshie8/r2hjr3js/4/

- mathmath333

nice

- mathmath333

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}

- ganeshie8

right you should get 7C4 = 35 combinations

- ganeshie8

for permutations, notice that each of that combination above can be permuted in 4! ways
so number of permutations = 35*4!

- mathmath333

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"

- amilapsn

yep... If you aren't allowed to use the same number twice.

- mathmath333

ok thnx

- mathmath333

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

##### 1 Attachment

- ganeshie8

you're not alone, everybody is facing that issue
@Astrophysics is there any fix yet ?

- amilapsn

me too...

- mathmath333

lol how can so many people tolerate this

- mathmath333

previuosly it happened only once in an hour but recently it get to on every long text i type

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