How many words can be formed by the letters of the word EXAMINATION taken all together?

Misc 4 - Chapter 7 Class 11 Permutations and Combinations (Term 2)

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Misc 4 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E? In dictionary, words are given alphabetically We need to find words starting before E i.e. starting with A,B,C or D In EXAMINATION, there is no B,C or D ,hence words should start with A Words in the list before the word starting with E = Words starting with letter A Words starting with letter A Since letter are repeating, Hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Remaining letters = n = 10 Since, 2I, 2N ∴ p1 = 2, p2 = 2 Words starting with letter A Since letter are repeating, Hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Remaining letters = n = 10 Since, 2I, 2N ∴ p1 = 2, p2 = 2 Number of words formed by there letters = 𝑛!/𝑝1!𝑝2! = 10!/2!2! = 907200 Thus, Words in the list before the word starting with E = Words starting with letter A = 907200

In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times and all the other letters appear only once.

The words that will be listed before the words starting with E in a dictionary will be the words that start with A only.

Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.

Since there are 2 Is and 2 Ns in the remaining 10 letters,

Number of words starting with A = `(10!)/(2!2!)` = 907200

Thus, the required numbers of words is 907200.

how many words can be formed using all letters in the word EXAMINATION

Assuming any sequence of letters is a word, how many words can we form in such a way that the first two letters are different consonants while the last two letters are vowels?

Assuming each letter is used only as many times as it occurs in the word.

First count the ways to place vowels and consonants, without considering identity or order within their type.

In addition to the two places on each ends, reserved for two consonants and two vowels respectively, there are seven places in the middle of the arrangement, which can hold three consonants and four vowels in $\frac{7!}{3!4!}$ ways.

$$\oplus\oplus\,\underbrace{\oplus\oplus\oplus\otimes\otimes\otimes\otimes}_{\frac{7!}{3!4!}\text{ permutations}}\,\otimes\otimes$$

Next count the ways to fill those places.

The consonants: XMTNN, can be arranged in $\frac{5!}{2!}-3!$ distinct permutations such that the pair of N do not simultaneously occupy the first two positions.

The vowels: EOAAII can be arranged in $\frac{6!}{2!2!}$ distinct permutations. If the vowels in the last two places have to be distinct this would be $\frac{6!}{2!2!}-2\times\frac{4!}{2!}$ permutations.

Thus there are $\frac{7!}{4!3!}\!\!\left(\frac{5!}{2!}-3!\right)\!\!\left(\frac{6!}{2!2!}\right)$ ways to arrange the letters as specified. ($340,200$)

There are $\frac{7!}{4!3!}\!\!\left(\frac{5!}{2!}-3!\right)\!\!\left(\frac{6!}{2!2!}-2\times\frac{4!}{2!}\right)$ ways if the vowels in the last two places have to be distinct. ($294,840$)

How many words can be formed by the letters of the word EXAMINATION taken all together how many words of them have no vowels coming together?

How many different ways can the letters of the word EXAMINATION?

How many ways 4 letters can be selected from the word EXAMINATION?

How many vowels are there in EXAMINATION?