a easy probability question, i think..... (1 Viewer)

[Hikari Clover]

5 letters r chosen from the letters of the word CRICKET. these five letters r then placed alongside one another to form a five letter arrangement. find the number of distinct five letter arrangements which r possible....

i m not sure with my answer:wave:

 

[victorheaven]

i got 1260.. is it right?

 

[kony]

consider case 1, where there is no C in the group of 5:

we have the letters RIKET to form words of 5 letters long. therefore, 5P5 = 120


consider case 2, where there is exactly 1 C:

to arrange the remaining 4 spots, we have 5 choices (again RIKET), so 5P4.
to arrange the 1 C, it's 5C1

therefore, 5P4 x 5C1 = 600


consider case 3, where there is exactly 2 Cs:

to arrange the remaining 3 spots, we have again the 5 choices, so 5P3.
to arrange the 2 Cs, we have 5C2

5P3 x 5C2 = 600


adding, 120+600+600 = 1320.

 

[Hikari Clover]

回复: Re: a easy probability question, i think.....

thanks , i got 1320 as well ~~