What is the number of ways in which the letters of the word table can be arranged so that the vowels occupy even places?

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What is the number of ways in which the letters of the word table can be arranged so that the vowels occupy even places?

Uh-Oh! That’s all you get for now.

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1. Four men and four women are to be seated alternatley a) in a row b) at a round table. In how many ways can this be done? 2. Hiow many nuymbers greatewr than 4000 can be formed from 3,5,7,8,9? (REPITION NOT ALLOWED)\ 3. In how many ways can the letters of the word PERMUTE be arranged if a) first and last posiitons occupied by consonants b) vowels and consonants occupy alternate places 4. How many numbners 0f 7 digits can be formed from the digits 1,2,3,4,5,6,,7 if: a)numner begins 1 and ends with 2 b)there are not more than 2 digits between 1 and 2? 5. In how many wasy can 5 diff math books, 4 diff physics books and 2 diff chemisty books be arranged on a shelf if the books in each subject are together? 6. In how many ways can 3 men, 3 women, 3 boys be arraniged in a row if 3 boys to remain togeterr 7. How amny ways can letyers of PRINICPLE be arranged.

8. how manny ways can letters of PRECISION arranged? How many of these arrangements fo vowels occupy even places?

1a) 8! = 40320 b) 7! = 5040 3a) 4x3x5! = 1440 b) 4x3x3x2x2x1x1 = 144 4a) 5! = 120 b) 5! x 30 = 3600 7. PRINCIPLE

9! / 2!x2! = 909720

What is the number of ways in which the letters of the word table can be arranged so that the vowels occupy even places?

1a) If they're steated alternatively, then it'd be 4! x 4!, then by 2!.. So 1152 1b) Since a table is (n-1)!, it'd be 4! x 3! .. So 144 2) Do 4 digit numbers first, so it'd be 4 x 4 x 3 x 2, Then do 5 digit numbers, so it'd be 5 x 4 x 3 x 2 x 1, then add to get 216. 3a) 4 consonants, 3 vowels... so 4 x 3 x 5! = 1440, then you divide by 2! as there are two E's, equalling 720 3b) 4! x 3! / 2! = 72 4a) 1 x 1 x 5! = 120 4b) This takes a while, you gotta consider if there are no digits, 1 digits or 2 digits between them. So for no digits = (2!) x 6! = 1440 Then for 1 digit in between = (1 x 1 x 5 x 2!) x 5! = 1200 Then for 2 digits in between = (1 x 5 x 4 x 1 x 2!) x 4! = 960 Add the three to get 3600 5) Maths (5!) x Physics (4!) x Chem (2!), then x 3! to arrange different subjects to get 34560. 6) Three boys = 3!. then x by 7! = 30240 7) 9!, then divide by 2! for two I's and 2! for two P's = 90720 8) 9! divided by 2! =181440

Then for evenly placed vowels, 5! x 4! / 2! = 1440

Fark, I suck at permutations

What is the number of ways in which the letters of the word table can be arranged so that the vowels occupy even places?

Fark, I suck at permutations

Don't worry about it dude, many people suck at permutations and probability in general. There are numbers of people who perform exceptionally at maths, then falter in probability.

Answer

What is the number of ways in which the letters of the word table can be arranged so that the vowels occupy even places?
Verified

Hint: Here, use the rules of permutation as this is a question of arrangement. Keep in mind the positions of vowels i.e. vowels must be in second and fourth places. So first find the number of ways for consonants and then find the number of ways for vowels, and multiply both the results.

Complete step-by-step answer:

We are given the word DELHI, which has 5 letters.Also there are two vowels E and I.And it is given that all vowels should occupy only the even places.Even places in the word consisting of 5 letters means second and fourth places.Number of ways for two vowels to take second and fourth places (which are even places) can be given as ${}^2{P_2}$.[Permutation formula: According to permutation formula, if we have total n object and we have to choose p objects then it can be done in ${}^n{P_r}$ ways i.e. $\dfrac{{n!}}{{(n - r)!}}$ ]And also the number of ways to arrange the remaining 3 letters which are consonants (D, L, H) in three places (i.e. odd places) can be given as ${}^3{P_3}$Now, ${}^2{P_2} = \dfrac{{2!}}{{(2 - 2)!}} = \dfrac{{2!}}{{0!}} = 2$ [Since, 0! = 1]And ${}^3{P_3} = \dfrac{{3!}}{{(3 - 3)!}} = \dfrac{{3 \times 2}}{{0!}} = 6$ [Since, 0! = 1]In permutation ‘and’ means multiplication and ‘or’ means addition. Here both the above arrangement for vowels and for consonants will happen together, therefore, the two results will be multiplied.So, the total number of ways to arrange the word DELHI arranged, so that the vowels occupy only the even places is 2 × 6 = 12.Hence, the correct option is (C).

Note: In these types of questions simply apply the rules of permutation, with given conditions. Do not try to write all the possible results as it may become very large in quantity. Hit and trial method will not work for more questions.