7. There are 5 doors to a lecture room. In how many ways can a student enter the room through a door and leave the room by a different door?
Answers
Answer:
20
All possible combinations:
If the student entered through A, she could leave through B or C or D or E.
There are 4 possibilities of entering through A and leaving through a different door viz., AB, AC, AD, AE.
If the student entered through B, she could leave through A or C or D or E.
There are 4 possibilities of entering through B and leaving through a different door viz., BA, BC, BD, BE.
If the student entered through C, she could leave through A or B or D or E.
There are 4 possibilities of entering through C and leaving through a different door viz., CA, CB, CD, CE.
If the student entered through D, she could leave through A or B or C or E.
There are 4 possibilities of entering through D and leaving through a different door viz., DA, DB, DC, DE.
If the student entered through E, she could leave through A or B or C or D.
There are 4 possibilities of entering through E and leaving through a different door viz., EA, EB, EC, ED.
There are a total of 4 + 4 + 4 + 4 + 4 = 20 ways.
We could also arrive at the same answer by saying - there are 5 ways to enter AND 4 ways to leave.
PLS MARK BRAINLIEST
Answer:
No. of doors = 5
No of ways to enter is 5
No of ways to exit is 5
So No. of ways by which he can enter then exit the room is 5 * 5 = 25 - 5 = 20
Step-by-step explanation:
When you need to mix no. of events into one whole event multiply possibilities. Like in this case two individual events were mixed to create one. So we multiplied the possibilities of both to gain the possibilities of the whole. Subtract 1 for each door common to enetr and exit.
Proof:
If the doors are A,B,C,D,E and first letter for enter second for exit. Then possible ways are: AA,AB,AC,AD,AE,BA,BB,BC,BD,BE,CA,CB,CC,CD,CE,DA,DB,DC,DD,DE,EA,EB,EC,ED,EE. Count and you get 25 possible ways to enter and exit.
Subtract AA,BB,CC,DD,EE and its twenty.