Both of your answers are incorrect. Method 1: We arrange the consonants, then place the vowels in the spaces between and at the ends of the row. There are $4!$ ways of arranging the four distinct consonants B, G, L, R. For each such arrangement, there are five spaces in which we could place the vowels, three between successive consonants and two at the ends of the row. To separate the vowels, we must place the three vowels in three of these five spaces. Choose two of the five spaces for the As and one of the remaining three spaces for the E. Thus, the number of admissible arrangements is $$4!\binom{5}{2}\binom{3}{1} = 720$$ Method 2: We use the Inclusion-Exclusion Principle. There are seven letters in total, including two As, one B, one E, one G, one L, and one R. There are $\binom{7}{2}$ ways to choose two of the seven positions in the arrangement for the As. The remaining five letters can be placed in the remaining five positions in $5!$ ways. Hence, there are a total of $$\binom{7}{2}5! = \frac{7!}{2!5!} \cdot 5! = \frac{7!}{2!}$$ ways of arranging the letters of the word ALGEBRA. From these, we must subtract those arrangements in which one or more pairs of vowels are consecutive. A pair of vowels is consecutive: Two As are consecutive or an A and E are consecutive. Two As are consecutive: We have six objects to arrange. They are AA, B, E, G, L, R. Since they are distinct, they can be arranged in $6!$ ways. An A and an E are consecutive: We have six objects to arrange. They are B, E, G, L, R, and a block containing an A and an E. The six objects are distinct, so they can be arranged in $6!$ ways. The A and E can be arranged within the block in $2!$ ways. Hence, there are $6!2!$ such arrangements. If we subtract the number of arrangements in which a pair of vowels is consecutive from the total, we will have subtracted each arrangement in which two pairs of vowels are consecutive twice, once for each way we could have designated one of those pairs as the pair of consecutive vowels. Since we only want to subtract such arrangements once, we must add them to the total. Two pairs of consecutive vowels: Since there are only three vowels, this can only occur if the three vowels are consecutive. We have five objects to arrange, B, G, L, R, and the block of three vowels. Since the objects are distinct, they can be arranged in $5!$ ways. The three vowels can be arranged in three ways: AAE, AEA, EAA. Hence, the number of such arrangements is $5! \cdot 3$. By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\frac{7!}{2!} - 6! - 6!2! + 5! \cdot 3 = 720$$ How many ways can the letters of the word PROMISE be arranged so that the vowels always come together? from cheatatmathhomework
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The letters of the word PROMISE are arranged so that no two of the vow [#permalink]
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Question Stats: Hide Show timer StatisticsThe letters of the word PROMISE are arranged so that no two of the vowels should come together. Find total number of arrangements. A. 7B. 49 C. 1.440 D. 1,898 E. 4,320 _________________
The letters of the word PROMISE are arranged so that no two of the vow [#permalink] Really Good Question i learned it the hard way , hope my solution helps __C__C__C__C__ Here is the arrangement of 4 Consonant such that between any two Consonant at most one Vowel can come but since the Vowels can also appear before the leftmost Consonant or after the rightmost Consonant so it gives us 5 places available for Vowels(represented by “___”)But since we require only 3 places due to 3 vowels to be arranged so we will select the 3 places out of 5 in 5C3 ways = 10 waysIn the given arrangement the Consonants can be arranged in 4! Ways at the selected 4 places and,All the vowels can also be arranged among themselves in 3! Waysso Total ways to arrange the letters as per desired condition = 10*3!*4! = 1440Answer: Option C _________________
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Re: The letters of the word PROMISE are arranged so that no two of the vow [#permalink] This can also be solve as an overlapping set problem - 7!(total arrangements without restrictions) - 3.2.6!(arrangements when any 2 of three are together) + 5!.6 (removing one duplicate overlapping of all three together from previous as this will happen twice) = 1440. Ans C.
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Re: The letters of the word PROMISE are arranged so that no two of the vow [#permalink]
Bunuel wrote: The letters of the word PROMISE are arranged so that no two of the vowels should come together. Find total number of arrangements. A. 7B. 49 C. 1.440 D. 1,898 E. 4,320 Combine vowels together so we are left with 4 places for Consonants and 1 for vowels out of 7 we will have now 4+1 = 5 ways and since 3 vowels are given then the combination to arrange this array is 5c3Now since no 2 vowels are to be together , so no. of ways to arrange C in 4 places 4 ! and vowels 3! = total arrangements hence possible = 5c3*4!*3! = 1440 option C
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Re: The letters of the word PROMISE are arranged so that no two of the vow [#permalink] Let '__' denote the possible locations of the vowels in the resultant word.Since no two vowels can be together, there must be at least one consonant between them. __C1__C2__C3__C4__Number of ways in which consonants can rearrange among themselves = 4! = 24 waysNow, we have 5 different spots that can accommodate vowels.Number of vowels in PROMISE is 3.We have 5 spots; we need 3. Does order of appearance of vowels matter? Of course! Since we have to count possible arrangements here, order does matter.Therefore number of ways of choosing 3 spots from a pool of 5 (for our vowels) is 5P3 = 5!/2! = 60.Therefore total number of ways in which the acceptable arrangements can be achieved = 24*60 = 1440 ways _________________
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Re: The letters of the word PROMISE are arranged so that no two of the vow [#permalink]
Bunuel wrote: The letters of the word PROMISE are arranged so that no two of the vowels should come together. Find total number of arrangements. A. 7B. 49 C. 1.440 D. 1,898 E. 4,320 I have two solutions I wanted to share. 1. Using straight calculationWe know the set of consonants = C = {P, R, M, S} and set of vowels = V = {O, I, E}.Since no vowels should come together, all the consonants are placed in between them, or they form the start and end of the word. Hence the format for the acceptable answer is : _ C _ C _ C _ C _ Now, there are 4 consonants and we are using all the consonants. The way 4 consonants can be arranged when using all of them is 4!Talking about the vowels, we have 3 vowels and 5 place for them. So we have to pick 3 positions where we can put those 3 vowels. No. of ways we can pick 3 places from 5 available is 5C3.Once we have picked 3 places, we need to arrange three vowels in the 3 places picked. We can do that in 3! ways.Hence total no. of ways is : 4 ! * 5C3 * 3 ! = 14402. Using negationno. of ways so that no two of the vowels are together = total no of ways - no. of ways so all 3 vowels are together - no. of ways so 2 of the vowels are together I know 2nd option becomes cumbersome but was a great exercise in thinking through different way to do the problem.
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Re: The letters of the word PROMISE are arranged so that no two of the vow [#permalink] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: The letters of the word PROMISE are arranged so that no two of the vow [#permalink] 13 Jun 2021, 09:18 |
How many ways can the letters of the word promise be arranged so that no two vowels come 49 B 1440 C 798?
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