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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 14:41:58 IST
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How many arrangements can be made with all letters of the word "VENUS" such that the order of the vowels remains unaltered.-I cant understand what order means in this problem.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 14:57:10 IST
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maybe he wants dat posn. of vowels must not change...in this case no. of arrangements would be:- 3! = 6
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 16:48:47 IST
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here order of the vowels remains unaltered means order of vowel doesnot take place ,In VENUS , E is on the second position and U is on fourth position , so order of vowels are 2 and 4 .. for solving the question . we arrange only V, N, S. So, Ans :- 3P3 = 6 . is it clear now.then rate me.
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I like to solve the tough problems, because nothing is tough.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 17:09:43 IST
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the word order means that the position of E and U in the word VENUS will remain same. in that case, the no. of words will be 3!*2!=6*2=12.
pls rate if the answer is correct..
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Agar aap 90 baar paap karoge to keval 45 baar hi pakde jaoge......batao
kyu????* *
b'cos.... sin90=cot 45 |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 17:25:06 IST
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no guys.........here order means that U should always come after E. now there are 4 possibilities. case I : when E occupies the first position, no. of words = 4! = 24 case II : when E occupies 2nd pos., no of words = 3*3*2*1= 18 case III : when E occupies the 3rd pos. , no of words = 3*2*2*1 = 12 case IV : when E occupies the 4th position, no of words =3! = 6
So, total no. of words possible = 24 + 18 + 12 + 6 = 60
RATE ME IF ITS CORRECT
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 17:49:35 IST
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the answer given by abhi3_raj is correct.
but the calculation can be done in an easier way.
there are 5! total possible cases. in half of these cases , E will be after U (e.g. NESVU ) and in the other half , U will be after E ( i.e. NUSVE)
so required no. of cases is 5! / 2 = 120 /2 = 60
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 21:41:41 IST
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As per the question the word "VENUS" can be arranged in : 3 factorial ways. soln:- The reason behind it is the vowels 'E' & 'U' must not change their orders,so the order of 'E' & 'U' remains same.So the left over three places can be filled in 3 factorial ways i.e = 6ways.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 09:56:58 IST
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In these all answer of this question which one is correct . please reply , anyone with full confidence. thanks.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 10:01:52 IST
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order of vowels does not change does not mean that positions should remain te same....... the order i.e the same order that appears in the word must remain the same......... soif the order of vowels in the word must remain same in EARTH. it means that A should come after E always (even if the positions of A and E are not same) soln posted by abhi3raj is hence right........
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 10:06:59 IST
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i got a formula for such problems in one of our books
the number of permutations of n different things taken r at a time in which x particular things always occur in assigned order is n-xP r-x (r-x+1)
in the given question, we have to permutate 5 letters taking 5 at a time(i.e, all at a time) in which 2 things always occur at the same order. so puttinh them in formula,
no of reqd permutations= 5-2P 5-2 (5-2+1)= 24
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it is not important where u stand, but in which direction u are moving |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 10:08:52 IST
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in this problem order means that E always comes before U
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it is not important where u stand, but in which direction u are moving |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 21:46:08 IST
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Another method is as follows place E & U first _E _ U_ now three letters remaining are V,N,S so for V three places are available (before E ,b/w E & U , or after U so V can be placed in 3 ways now after V is placed , N can be placed in 4 ways(as an extra gap will be created , b/w V & E or b/w U & V) similarly after placing N , S can be placed in 5 ways so total ways are 5 * 4 * 3 = 60 this method is called placing technique and is useful when arrangement cases become too many. for eg: if a question comes when there are five vowels and order shoud remain same then this method should be used.
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