How many ways $INSTRUCTOR$ can be arranged such that it has two consecutive vowels?(Three consecutive vowels are not counted i.e $V1V2V3$ is not counted). Show I was trying to solve this problem in a different way. But I got a mismatch in the answer. Following is my solution INSTRUCTOR has three vowels $I,U$ and $O$. Lets arrange the three vowels like this _ V1V2 _ V3 _ The above representation shows V1,V2,V3 as vowels and "_" as consonants. We have total 7 consonants which can fill all these three holes(refereeing to "_"). There must be at least one consonant between V1V2 and V3, else three vowels will come together. It can be written like this: $X_1+X_2+X_3=7$ $X_1\ge0,X_2\ge1,X_3\ge0$ This will give result $C(8,2)=28$ We can arrange these 3 vowels in $3!$ ways. Again we can also interchange the position of $V_1,V_2$ and $V_3$ like this. _V3_V1V2_ So we have total $(28)(3!)(2)$ ways of arranging three vowels. Now the other 7 consonants can be arranged in $\frac{(7!)}{(2!)(2!)}$ ways. So answer should be $$\frac{(28)(3!)(2)(7!)}{(2!)(2!)}$$ Given answer $$\frac{(64)(3!)(7!)}{(2!)(2!)}$$ What's wrong here? Help appreciated :) $\begingroup$ I went on to proceed with this problem by subtracting the number of arrangements in which all vowels were together, from the total number of arrangement which proved time consuming. An alternative solution, given by the textbook was what they called the gap method. They find the number of ways of arranging $6$ consonants. "Then they say that there are $7$ gaps remaining (how!!!!!)" in which the remaining $6$ vowels are arranged in $\frac{7p6}{3!2!}$. The answer is $151200$
asked Jan 19, 2017 at 11:12
$\endgroup$ 1 $\begingroup$ The vowels are: AEEEII. The non-vowels are: DMNRTT. The number of ways to arrange the vowels is $\frac{(1+3+2)!}{1!\times3!\times2!}=60$. The number of ways to arrange the non-vowels is $\frac{(1+1+1+1+2)!}{1!\times1!\times1!\times1!\times2!}=360$. The number of ways to choose slots for the $6$ vowels between the $6$ non-vowels is $\binom{6+1}{6}=7$. So the number of ways to arrange this word under the given restriction is $60\cdot360\cdot7=151200$. answered Jan 19, 2017 at 11:57
barak manosbarak manos 42.4k8 gold badges53 silver badges129 bronze badges $\endgroup$ 2 $\begingroup$ Take one easy example - Suppose you have word UGLIER . In how many ways arranged so that vowels never comes together. We have 5 letters and we first fill consonants (C) alternatively in them. _ , C , _ , C , _ , C Now we have to fill vowels. But you can notice that we can create one place at last and delete starting place. So we have 4 places to fill 3 vowels. answered Jan 19, 2017 at 11:54
Kanwaljit SinghKanwaljit Singh 8,5341 gold badge8 silver badges17 bronze badges $\endgroup$ 2 $\begingroup$ how may gaps are there between the consonants..$6$ consonants so for vowels there are $7$ places.first arrange the consonants in $(6!/2)$ ways.now choose any 6 places from the 7 and arrange the 6 vowels that can be done in ($7\choose 6$$×6!)/(2!×3!)$ and finally multiply both the arrangements answered Jan 19, 2017 at 11:53
UpstartUpstart 2,58911 silver badges25 bronze badges $\endgroup$ 6 How many ways vowels never come together?number of arrangements in which the vowels do not come together =5040−1440=3600 ways.
How many different ways can the letters of a be arranged so that vowels always come together?In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together? A. 1440.
How many words can be formed of the letter topology in which two vowels are never together?Hence the total number of ways when the two vowels never come together by fundamental theorem is 360×420=151200 by (1), (2)
How many ways can vowels be arranged?The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.
|
