How many 4 letter words can be made from the word EXAMINATION
@hartnn , @ganeshie8
@wio
exam ,mint,next,none
a lot.
First split up int distinct groups of letters.
YA , i know the beginning process and everything , the answer is 2454 @wio , just give me the detailed explanation
the three cases , etc
EXAMINATION E X M T O I I N N A A First classify based on number of letters
okay , then
First case is all distinct letters, which is \(^8P_4\)
Second case is 2 distinct letters and then 1 repeated letter
ya , so C(3,1) * C(7,2) * 4factorial ?
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.
Actually I think I might be a bit wrong on the choosing 7 distinct letters part.
ya , that step is incorrect
:P
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
the answer is also same
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?
ya
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.
why not just C(3,2) * 4fact/2fact*2fact
BTW , Congrats on 99 , last time i saw u were 92
It's the same thing
HMM , OK
and then add , ryte
In fact I definitely should have just said \(^3C_2\)
?
:D
Each case has its steps multiplied, because they are sequential (independent steps) The cases themselves are added because they are simultaneous (mutually exclusive cases)
ya , that's what i meant :D
thanks bro :D
Cool, make sure it works out to the right answer.
umm , how do i give a testimonial ?
ya , it does
hover over name, then become fan, then testimonial button appears
You just made me 90 in problem solving.
HAHA , it's the least i could do
What answer did you get?
what the actual?? aren't you falsely assuming that every 4-letter combination is a word?
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