Perms (1 Viewer)

[cutemouse]

The letters of the word CALCULUS are arranged in a row
(i) How many different arrangements are possible? (I can do this) = 5040
(ii) If one of these arrangements is selected at random, what is the probability that it begins with 'U' and ends in 'U'?

I can do (ii) using 2U level probability (ie. 2/8 * 1/7), but surely there must a be a 3U way that incorperates the answer in (i)?

Thanks

 

[lychnobity]

The betters of the word CALCULUS are arranged in a row
(i) How many different arrangements are possible? (I can do this) = 5040
(ii) If one of these arrangements is selected at random, what is the probability that it begins with 'U' and ends in 'U'?

I can do (ii) using 2U level probability (ie. 2/8 * 1/7), but surely there must a be a 3U way that incorperates the answer in (i)?

Thanks

ii) You're 2u level method is wrong imo

It should be something like:

(6!/(3!2!))/5040

 

[scardizzle]

ii) You're 2u level method is wrong imo

It should be something like:

(6!/(3!2!))/5040

shouldn't it be (6!/(2!2!))/5040 since there are 2 C's and 2 L's where are you getting the 3! from?

 

[Timothy.Siu]

The betters of the word CALCULUS are arranged in a row
(i) How many different arrangements are possible? (I can do this) = 5040
(ii) If one of these arrangements is selected at random, what is the probability that it begins with 'U' and ends in 'U'?

I can do (ii) using 2U level probability (ie. 2/8 * 1/7), but surely there must a be a 3U way that incorperates the answer in (i)?

Thanks

i)8!/2!2!2!
ii)(6!/2!2!)/(8!/2!2!2!)

 

[Timothy.Siu]

ii) You're 2u level method is wrong imo

It should be something like:

(6!/(3!2!))/5040

its right.
its basically the same thing.
his way, you're doing the probability of having a word with first U and last U.
2/8 x 1/7. its definitely right.

 

[cutemouse]

ii)(6!/2!2!)/(8!/2!2!2!)

How'd ya do that?

 

[mchew92]

Hey min?
Its pretty much CALCULUS ignoring the Us

so its 6!/(2!2!) / 5040

where 8!/2!2!2!= 5040