Rooms for Guests ;)

there are 3 guests ,, each guest should have a separate room .. for this first guest has 6 choices of selecting his room.. he selects one room for him then second guest has 5 choices and he selects one room .. third guest has 4 choices... therefore 6 * 5 * 4=120 hence solved.. 120 is the ans.. :)

There are 6 spare rooms. 3 rooms can be selected of the 6 by 6C3=20.

The 3 guests can be arranged in 3 rooms by 3! ways=6

Total number of ways= 20x6= 120.

The first guest can be accommodated in any of the 6 different rooms. The second guest can be accommodated in any of the 5 remaining rooms. The third guest can be accommodated in any of the 4 remaining rooms. Therefore, total number of ways in which the guests can be accommodated= 6x5x4 = 120 :D :D

6!/(6-3)!=120

(6!)/(3!)?

6!/3! . you get this how. there are 6 choice. so 6! but 3 geusts so6!/3!

6 options...3 choices => \frac { 6! }{ 3! }

3! x 6 C 3 or 6 P 3 = 6 X 5 X 4 = 120

First: 6 ways;

Second: 5 ways;

Third: 4 ways;

Consecutive events: AND multiplication.

Using permutation, 6P3 = 120.

it's like a simple problem: how many do you have to write a 3-number number from 6 numbers. and the answer will be: 6x5x4 = 120

$C^{6}_{1} * C^{5}_{1} * C^{4}_{1}$

is permutations of 6 over 3. 6 factorial over (6-3) factorial equals to120