Solution of this questions with explanation (PEMUTATION)

[[Announcements] [Introduce Yourself] [ICT] [Prasekolah] [LINUS] [KSSR] [UPSR] [PMR] [SPM] [STPM] [School] [Classroom] [Clubs & Uniformed Bodies] [JPN] [SPA] [SICT] [SPP] [SPS] [SPPK] [SPSPK] [SPI] [SPsK] [JNJK] [PPD] [Program] [Computer & Internet] [Let's Hear It From You] [Membergroup]]

[Announcements] [Introduce Yourself] [ICT] [Prasekolah] [LINUS] [KSSR] [UPSR] [PMR] [SPM] [STPM] [School] [Classroom] [Clubs & Uniformed Bodies] [JPN] [SPA] [SICT] [SPP] [SPS] [SPPK] [SPSPK] [SPI] [SPsK] [JNJK] [PPD] [Program] [Computer & Internet] [Let's Hear It From You] [Membergroup]

[Yap Yee Soon]

Can any experience teacher give the full solution and the explanation ?

thanks


1.   In how many ways can all the letters in the word  ‘SECONDARY’ be arranged without repetition such that
(a)   the vowels placed adjacent to each other ?
(b)   the vowels separated from each other ?

2.   In how many ways can all the letters in the word  ‘RESPONSIBILITIES’ be arranged without repetition such that
(a)   the vowels placed adjacent to each other ?
(b)   the vowels separated from each other ?

[Teh Guan Leong]

 

[Teh Guan Leong]

Let me try the question

1
a)
the vowels placed next to each other = 3P3
then the vowels bounded up as 1 unit then arrange again with the consonants left = 7P7

Total arrangements = 3P3 x 7P7 = 30240
b)
when we find total arrangements where vowels seperated from each other, fastest way is we find total arrangements without any conditions minus total arrangements where the vowels are arranged together

Total arrangements = 9P9 - 30240 = 332640

No 2 is a bit different because there are some repeating letters. It can only be found at Form 6 syllabus. However if u wanted the working,
a)
Arrangements where the vowels are together, 7P7/(2!4!)
All the vowels bounded up as 1 unit and arranged together with the consonants left = 10P10/3!

Total arrangements = 7P7/(2!4!) x 10P10/3!

b)
when we find total arrangements where vowels seperated from each other, fastest way is we find total arrangements without any conditions minus total arrangements where the vowels are arranged together

Total arrangements = 16P16/(2!4!3!) - 63504000

Correct me if I'm not right  

[Wong Tok Hua]

I think for 1(b), we need to minus off the case where two vowels are next to each other. The number of arrangement with 2 vowels next to each other is 2!x8!x3

For 2(b), we need to arrange the consonants as follows:
                _ R _ S _ P _ N _ S _ B _ L _ T _ S _
The vowels must be arranged in those 10 spaces in between consonants. That means we have 10P7/(3!2!) ways to arrange the vowels.
But we have 9!/3! ways to arrange the consonants.
So, the total number of arrangements should be 10P7/(3!2!)x9!/3!.

[Yap Yee Soon]

Thank you Mr Wong and Mr Teh, regarding to 1b, i was confuse cuz some ref books answer are without (minus) the the 2 vowels like Mr Teh's answer But some ref books are with (minus) the  next to each other like you (Mr Wong).

Anyone can tell me which is the correct answer and how to solve 1b with the easiest way?

thank you

[Lim Yu Teong]

The answer for 1b is 7P3 X 6! .

[Yap Yee Soon]

mr Lim, can u explain why??

[Voon Nyuk Fung]

For Q1b, considering the spaces between the consonants (5 in between and 2 at the ends, total 7) thus the vowels can be arranged (separated) in 7P3 ways. There are 6! arrangements between the consonants, thus the total is 7P3 X 6!.

[Wong Tok Hua]

That right. Thanks, Mr. Lim. The method is the same as 2b.
BTW, the number of arrangement with 2 vocals side by side should be 2!x7!x6x3 = 181440.

Advanced search