Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (anonymous):

How many 4 letter words can be made from the word EXAMINATION

OpenStudy (anonymous):

@hartnn , @ganeshie8

OpenStudy (anonymous):

@wio

OpenStudy (anonymous):

exam ,mint,next,none

OpenStudy (anonymous):

a lot.

OpenStudy (anonymous):

First split up int distinct groups of letters.

OpenStudy (anonymous):

YA , i know the beginning process and everything , the answer is 2454 @wio , just give me the detailed explanation

OpenStudy (anonymous):

the three cases , etc

OpenStudy (anonymous):

EXAMINATION E X M T O I I N N A A First classify based on number of letters

OpenStudy (anonymous):

okay , then

OpenStudy (anonymous):

First case is all distinct letters, which is \(^8P_4\)

OpenStudy (anonymous):

Second case is 2 distinct letters and then 1 repeated letter

OpenStudy (anonymous):

ya , so C(3,1) * C(7,2) * 4factorial ?

OpenStudy (anonymous):

I think first you choose 7 distinct letters \(^8C_7\) then you chose 1 repeated letter \(^3C_1\). Assign the distinct letters a position \(^4C_3\) and let the let the repeated letters fall in to their place.

OpenStudy (anonymous):

Actually I think I might be a bit wrong on the choosing 7 distinct letters part.

OpenStudy (anonymous):

ya , that step is incorrect

OpenStudy (anonymous):

:P

OpenStudy (anonymous):

huh , had this as a 6 marker on my test today , apparently my teacher had given the same question with "MATHEMATICS" and the students mugged it up

OpenStudy (anonymous):

the answer is also same

OpenStudy (anonymous):

First choose the repeated letter \(^3C_1\) and then choose the two distinct letters \(^7C_2\) and then assign them sports \(^4C_3\) That would make sense, right?

OpenStudy (anonymous):

ya

OpenStudy (anonymous):

For the final step, we would do both 2 repeated letters. So \(^3C_1\) and \(^2C_1\) gets us the two letters we want. Assign the first two distinct letters with \(^4C_2\) Assign the first repeated letter with \(^2C_1\), the last letter falls into place.

OpenStudy (anonymous):

why not just C(3,2) * 4fact/2fact*2fact

OpenStudy (anonymous):

BTW , Congrats on 99 , last time i saw u were 92

OpenStudy (anonymous):

It's the same thing

OpenStudy (anonymous):

HMM , OK

OpenStudy (anonymous):

and then add , ryte

OpenStudy (anonymous):

In fact I definitely should have just said \(^3C_2\)

OpenStudy (anonymous):

?

OpenStudy (anonymous):

:D

OpenStudy (anonymous):

Each case has its steps multiplied, because they are sequential (independent steps) The cases themselves are added because they are simultaneous (mutually exclusive cases)

OpenStudy (anonymous):

ya , that's what i meant :D

OpenStudy (anonymous):

thanks bro :D

OpenStudy (anonymous):

Cool, make sure it works out to the right answer.

OpenStudy (anonymous):

umm , how do i give a testimonial ?

OpenStudy (anonymous):

ya , it does

OpenStudy (anonymous):

hover over name, then become fan, then testimonial button appears

OpenStudy (anonymous):

You just made me 90 in problem solving.

OpenStudy (anonymous):

HAHA , it's the least i could do

OpenStudy (anonymous):

What answer did you get?

OpenStudy (anonymous):

what the actual?? aren't you falsely assuming that every 4-letter combination is a word?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!