Q. permutation/combination (2 Viewers)

[freaking_out]

the following question actually is from a 3u trial paper:

"All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contains all the vowels alphabetically in order but separated by at least one consonant."

 

[die_imortales]

All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contains all the vowels alphabetically in order but separated by at least one consonant

first u put in the 7 consonants(C) in 7! ways so u have:

/C/C/C/C/C/C/C/

where the /'s are the slots where the vowels can go....

then u just have to put A,E,E,U in that order in 7 spots

which would be 7P4 mutliplied by 1/4 to make sure A is first... and then 1/3 to make sure U is last and then 2 because the E's can be swapped which gives:

7! * 7P4 * 1/4 * 1/3* 2 = 705600.... mite be rite... if it is i'm cheerin.. if not then soz for wastin ur time

 

[freaking_out]

Originally posted by die_imortales
All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contains all the vowels alphabetically in order but separated by at least one consonant

first u put in the 7 consonants(C) in 7! ways so u have:

/C/C/C/C/C/C/C/

where the /'s are the slots where the vowels can go....

then u just have to put A,E,E,U in that order in 7 spots

which would be 7P4 mutliplied by 1/4 to make sure A is first... and then 1/3 to make sure U is last and then 2 because the E's can be swapped which gives:

7! * 7P4 * 1/4 * 1/3* 2 = 705600.... mite be rite... if it is i'm cheerin.. if not then soz for wastin ur time

i don't know, there are 7 consonants, but there are 2 "C"s, so u can't say 7!

 

[die_imortales]

soz didnt see that... well then its 7! divided by 2 because of the two C's
so divide answer by 2

 

[freaking_out]

Originally posted by die_imortales
soz didnt see that... well then its 7! divided by 2 because of the two C's
so divide answer by 2

well also the question says "atleast 1 consonant b/w the vowels" so u can have 2 consonants b/w the vowels, for which your answer does not account for.

 

[die_imortales]

yeah it does...
if u put the vowels in any 1 of those slots then u could have any number of consonants in between

 

[underthesun]

Wat iz de enser from de bek?

 

[die_imortales]

personally i would hav like to know that too....

 

[-=«MÄLÅÇhïtÊ»=-]

There are 7 consonants with 2 pairs of repeats, so they can be arranged in 7!/(2!2!) ways.

The vowels are in alpabetical order, so no need to permutate, juz gotta work out how many ways u can fit into spots separated by consonants. That's just 8C4. (I think this right...I'm not doing anything for 2 E's because 8C4 is only looking for places to put in the vowels. Theres only 1 way to arrange the vowels).

So answer's 7!/4 * 8C4 = 88200

 

[die_imortales]

oh yeh there are two repeating consonants arent ther... gotta be more observant.. otherwise the vowel part i had no idea about i'd just forgot... what ur sayin sounds rite tho the E's wont need to be muddled with cause u can just consider them two different vowels... well i guess i need to study some permutations... seein as i was about 620000 off...

 

[-=«MÄLÅÇhïtÊ»=-]

the way i see it, this ques has 3 steps

1) permutate consonants
2) find places to put in the vowels
3) permutate vowels, but no need coz only 1 way to stay in order

 

[freaking_out]

Originally posted by underthesun
Wat iz de enser from de bek?

me no have answer, it from ze ecsam peper....james ruse 2003.

 

[underthesun]

oh, meibi wi ken ask sam piple from jeims rus den

do those james ruse kids visit this forum, let alone browse the net or use computers? I mean, agricultural nerds.. j/k

 

[freaking_out]

Originally posted by underthesun
oh, meibi wi ken ask sam piple from jeims rus den

do those james ruse kids visit this forum, let alone browse the net or use computers? I mean, agricultural nerds.. j/k

i doubt any james ruse kids hand around ze forum...they're prolly too busy studying for their hsc now..

 

[OLDMAN]

Funny (or disconcerting) to see that some perm/comb ext1 trial questions are harder than some ext2 problems. I remember a similar CSSA trial (1997?) on the word EXTENSION, no two vowels together.

Again I'll be stepping on the backs of die_imortales and malachite's initial analyses to solve this one:

/C/C/C/C/C/C/C/

vowels EEUA, consonants NNCCMBR

8 slots for the vowels : 8P4/2!
7 slots for the consonants :7!/(2!2!)

Total no. of ways 8P4/2! * 7!/(2!2!) = 1,058,400

or, following malachite's lead :

8C4 ways of choosing 4 slots for the vowels,
4!/2! ways to place the 4 vowels in a particular 4 slots,
7!/(2!2!) ways of placing the consonants
Total no. of ways 8C4 * 4!/2! * 7!/(2!2!) = 1,058,400

 

[freaking_out]

Originally posted by OLDMAN
Funny (or disconcerting) to see that some perm/comb ext1 trial questions are harder than some ext2 problems. I remember a similar CSSA trial (1997?) on the word EXTENSION, no two vowels together....

8 slots for the vowels : 8P4/2!
7 slots for the consonants :7!/(2!2!)

Total no. of ways 8P4/2! * 7!/(2!2!) = 1,058,400

correct me if i'm wrong, but shouldn't the way u slot the vowels be: 8C4/2!, since u have to keep the vowels in alphabetical order?

also, yeah, i think that "EXTENSION" question was from the 2002 catholic paper.

 

[freaking_out]

another similiar question

since the following question looks similar i decided to put it in this thread:

if the letters of the word GUMTREE and the letters of the word KOALA are combined and arranged into a single twelve-letter word, in how many of these arrangements do the letters of KOALA appear in their correct order, but not necessarily together?

 

[OLDMAN]

___________________________________________________
quote : freakin_out
correct me if i'm wrong, but shouldn't the way u slot the vowels be: 8C4/2!, since u have to keep the vowels in alphabetical order?
___________________________________________________



oops! forgot the alphabetical order. divide by 12 to get malachite's answer, 88,200.

 

[Constip8edSkunk]

Originally posted by freaking_out
since the following question looks similar i decided to put it in this thread:

if the letters of the word GUMTREE and the letters of the word KOALA are combined and arranged into a single twelve-letter word, in how many of these arrangements do the letters of KOALA appear in their correct order, but not necessarily together?

7!/2! * (8C5+4*8C4+6*8C3+4*8C2+8C1) = 1995840

is this right?

 

[freaking_out]

Originally posted by Constip8edSkunk
7!/2! * (8C5+4*8C4+6*8C3+4*8C2+8C1) = 1995840

is this right?

ya, zats wat it zays, in ze back of ze book.

can u plz. ecsplain how?