How many different ways can the letters of the word statistics be arranged so that the vowels must always come together?

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    Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

    In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

    Permutation Formula

    In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

    nPr = (n!)/(n – r)!

    Here,

    n = group size, the total number of things in the group

    r = subset size, the number of things to be selected from the group

    Combination

    A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

    Combination Formula

    In combination r things are picked from a set of n things and where the order of picking does not matter.

    nCr = n!⁄((n-r)! r!)

    Here,

    n = Number of items in set

    r = Number of things picked from the group

    Solution:

    Vowels are: I,I,O,E

    If all the vowels must come together then treat all the vowels as one super letter, next note the letter ‘S’ repeats so we’d use

    7!/2! = 2520 

    Now count the ways the vowels in the super letter can be arranged, since there are 4 and 1 2-letter(I’i) repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

    = (7!/2! × 4!/2!) 

    = 2520(12)

    = 30240 ways

    Similar Questions

    Question 1: In how many ways can the letters be arranged so that all the vowels came together word is CORPORATION?

    Solution:

    Vowels are :- O,O,A,I,O

    If all the vowels must come together then treat all the vowels as one super letter, next note the R’r letter repeat so we’d use

    7!/2! = 2520

    Now count the ways the vowels in the super letter can be arranged, since there are 5 and 1 3-letter repeat the super letter of vowels would be arranged in 20 ways i.e., (5!/3!)

    = (7!/2! × 5!/3!)

    = 2520(20)

    = 50400 ways

    Question 2: In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together?

    Solution:

    Vowels are :- A,A,E,I

    Next, treat the block of vowels like a single letter, let’s just say V for vowel. So then we have MTHMTCSV – 8 letters, but 2 M’s and 2 T’s. So there are

    8!/2!2! = 10,080

    Now count the ways the vowels letter can be arranged, since there are 4 and 1 2-letter repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

    = (8!/2!2! × 4!/2!)

    = 10,080(12)

    = 120,960 ways

    Question 3: In How many ways the letters of the word RAINBOW be arranged in which vowels are never together?

    Solution:

    Vowels are :- A, I, O  

    Consonants are:- R, N, B, W.

    Arrange all the vowels in between the consonants so that they can not be together. There are 5 total places between the consonants. So, vowels can be organize in 5P3 ways and the four consonants can be organize in 4! ways.

    Therefore, the total arrangements are 5P3 * 4! = 60 * 24 = 1440

    This section covers permutations and combinations.

    Arranging Objects

    The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

    Example

    How many different ways can the letters P, Q, R, S be arranged?

    The answer is 4! = 24.

    This is because there are four spaces to be filled: _, _, _, _

    The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!

    • The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is:

    n!        .
    p! q! r! …

    Example

    In how many ways can the letters in the word: STATISTICS be arranged?

    There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are:

    10!=50 400
    3! 2! 3!

    Rings and Roundabouts

    • The number of ways of arranging n unlike objects in a ring when clockwise and anticlockwise arrangements are different is (n – 1)!

    When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)!

    Example

    Ten people go to a party. How many different ways can they be seated?

    Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440

    Combinations

    The number of ways of selecting r objects from n unlike objects is:

    How many different ways can the letters of the word statistics be arranged so that the vowels must always come together?

    Example

    There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls?

    10C3 =10!=10 × 9 × 8= 120
                 3! (10 – 3)!3 × 2 × 1

    Permutations

    A permutation is an ordered arrangement.

    • The number of ordered arrangements of r objects taken from n unlike objects is:

    nPr =       n!       .
              (n – r)!

    Example

    In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use.

    10P3 =10!
                7!

    = 720

    There are therefore 720 different ways of picking the top three goals.

    Probability

    The above facts can be used to help solve problems in probability.

    Example

    In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery?

    The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 .

    Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.

    Distinguishable Ways to Arrange the Word STATISTICS
    The below step by step work generated by the word permutations calculator shows how to find how many different ways can the letters of the word STATISTICS be arranged.

    Objective: Find how many distinguishable ways are there to order the letters in the word STATISTICS.

    Step by step workout:


    step 1 Address the formula, input parameters and values to find how many ways are there to order the letters STATISTICS.
    Formula:
    nPr =n!/(n1! n2! . . . nr!)

    Input parameters and values:

    Total number of letters in STATISTICS: n = 10 Distinct subsets: Subsets : S = 3; T = 3; A = 1; I = 2; C = 1; Subsets' count:

    n1(S) = 3, n2(T) = 3, n3(A) = 1, n4(I) = 2, n5(C) = 1

    step 2 Apply the values extracted from the word STATISTICS in the (nPr) permutations equation


    nPr = 10!/(3! 3! 1! 2! 1! )

    = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10/{(1 x 2 x 3) (1 x 2 x 3) (1) (1 x 2) (1)}

    = 3628800/72

    = 50400 nPr of word STATISTICS = 50400 Hence, The letters of the word STATISTICS can be arranged in 50400 distinct ways.

    Apart from the word STATISTICS, you may try different words with various lengths with or without repetition of letters to observe how it affects the nPr word permutation calculation to find how many ways the letters in the given word can be arranged.