Question

Counting question

Answer 1

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

Answer 2

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

Answer 3

is it 4!

Answer 4

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

Answer 5

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

How to implement this

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Simplicity @ganeshie8 :D lol

Answer 11

4! is true if you just have 4 chairs

Answer 12

....and only 4 people..

Answer 13

yes i just have 4 chairs

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

is there any website that shows permutations and combinations list

Answer 18

python has got the function permutations in itertools module.

Answer 19

ABC
ACB
ABD
ADB
ACD
ADC
BCD
BDC

Answer 20

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

Answer 21

nice

Answer 22

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}

Answer 23

right you should get 7C4 = 35 combinations

Answer 24

for permutations, notice that each of that combination above can be permuted in 4! ways

so number of permutations = 35*4!

Answer 25

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"

Answer 26

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

Answer 27

ok thnx

Answer 28

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

Answer 29

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

Answer 30

me too...

Answer 31

lol how can so many people tolerate this

Answer 32

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