5 people are waiting in line for their picture to be taken. How many different ways can the line formed?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Very well said.
The answer is 120. This is because in the first spot in the queue, there are 5 possible people who can go there. In the second spot in the queue, there are only 4 people who can go there, since one is already in the queue. This continues until the final spot in the queue, where only 1 possible person can go. Hence, the answer is 5×4×3×2×1, which equals 120.
We have to arrange 5 people in a line and this can be done by permuting 5 people taken all at a time, i.e, 5P5 = 5! = 120
This can be done by 5! ways so 5! = 5x4x3x2x1=120
5! = 120 as easy as that
The ways we form the line is the same as 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
5 ! = 120 solutions différentes.
Simply 5 factorial (5!).! 1st Line-> 5 people 2nd Line-> 5-1=4 3rd Line-> 5-2=3 4th Line-> 5-3=2 5th Line-> 5-4=1 Formula=5 4 3 2 1= 120 different ways.
'n' objects can be arranged in 'n' places in 'n !' ways. so, here 5 ! = 120 ways.
There are five(5) place to be occupied by five(5) person.
On the 1st place, the possibility the place will be filled is 5.
On the 2nd, the possibility the place will be filled is 4.
On the 3rd, the possibility the place will be filled is 3.
On the 4th, the possibility the place will be filled is 2.
leaving the 5th place to be occupied by the remaining 1.
the solution would be: (5)(4)(3)(2)(1) = 120
using probability/posibility of a person to takes place to other person wihout returning to his place!!!!!! 5 4 3 2 1=120
5 4 3 2 1=120 possible ways.
number of ways the line can be formed = 5!= 120
Let P1, P2.... P5 be the persons.It makes a lot of difference between the P1 standing ahead of P2 and viceversa. So the problem(in simpler words) is asking the number of permutations possible for 5 people and 5 places.
So, the answer is 5P5 or P (5,5) = 5! = 120
5c5
Log in to reply
No, since the people are being arranged in a line, arrangement and positions of the people in the line is important, so you cannot use combination as it only takes into view the selections of the people and does not count the different arrangements.
5C5=1 is wrong. Just common sense. :))
There are 5 choices for the 1st place.
There are 4 choices for the 2nd place.
This continues till we have 1 choice for the last place.
Implying, no. of choices to be 5!.
Hence the answer 120.
nice idea
Thus answer= 5!=5x4x3x2x1= 120 :)
Problem Loading...
Note Loading...
Set Loading...
There are 5 choices for the 1st place.
There are 4 choices for the 2nd place.
This continues till we have 1 choice for the last place.
Implying, no. of choices to be 5!.
Hence the answer 120.