How many ways can we arrange the letters in the word COMMITTEE

Answer

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Hint: Here, we are required to arrange the letters in the given word ‘FACTOR’. Thus, we will use Permutations to ‘arrange’ the letters keeping in mind that all the letters in the given word are unique. Thus, applying the formula and solving the factorial, we will be able to find the required ways of arrangement of letters of the given word.

Formula Used:

We will use the following formulas:1. ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ , where $n$ is the total number of letters and $r$ represents the number of letters to be arranged.2. $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.

Complete step-by-step answer:

In order to find the arrangement of the word ‘FACTOR’,First of all, we will observe that all the letters in this given word are unique and no word is the same or duplicate. Also, the number of letters in the word ‘FACTOR’ is 6.Therefore, we will use Permutations to ‘arrange’ the 6 letters of the given word.Thus, the formula is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$Where, $n$ is the total number of letters and $r$ represents the number of letters to be arranged, i.e. $6$ in each case.Thus, we get,${}^6{P_6} = \dfrac{{6!}}{{\left( {6 - 6} \right)!}} = \dfrac{{6!}}{{0!}} = 6!$Because, $0! = 1$Now, the formula of expanding factorial is $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.Hence, we get,$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 30 \times 24 = 720$

Therefore, we can arrange the letters in the word ‘FACTOR’ in 720 ways.
Thus, this is the required answer.

Note:

While solving this question, we should know the difference between permutations and combinations. Permutation is an act of arranging the numbers whereas combination is a method of selecting a group of numbers or elements in any order. Hence, Permutations and Combinations play a vital role to solve these types of questions. . Also, in order to answer this question, we should know that when we open a factorial then, we write it in the form of: $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$ as by factorial we mean that it a product of all the positive integers which are less than or equal to the given number but not less than 1.

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Probability

the word COMMITTEE has 9 letters so 9!= permutation of that word,b ut... since the letters M, T and E are repeated each twice, there will be words that are indestinguishable from each othe because one M is not different from the other M, and same for the T and E. -- So we need to back these words out. --- answer: 9!/(2!*2!*2!)= 45360

If it is possible to make a meaningful word with the first, the seventh, the ninth and the tenth letters of the word RECREATIONAL, using each letter only once, which of the following will be the third letter of the word? If more than one such word can be formed, give ‘X’ as the answer. If no such word can be formed, give ‘Z’ as the answer.

Answer & Explanation Answer: D) R

Explanation:

The first, the seventh, the ninth and the tenth letters of the word RECREATIONAL are R, T, O and N respectively. Meaningful word from these letters is only TORN. The third letter of the word is ‘R’.

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Answer: D) 43200

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

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