How many words can be formed with letters of EXAMINATION by taking all the letters at a time in how many of these two vowels do not come together?

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. ($294,840$)

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Exercise :: Permutation and Combination - General Questions

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2. 

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A. 360
B. 480
C. 720
D. 5040
E. None of these

Answer: Option C

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

How many words can be formed with letters of EXAMINATION by taking all the letters at a time in how many of these two vowels do not come together?
Required number of ways = (120 x 6) = 720.

Video Explanation: https://youtu.be/WCEF3iW3H2c

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Exercise :: Permutation and Combination - General Questions

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7. 

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

How many words can be formed with letters of EXAMINATION by taking all the letters at a time in how many of these two vowels do not come together?
Required number of numbers = (1 x 5 x 4) = 20.

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Exercise :: Permutation and Combination - General Questions

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13. 

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A. 10080
B. 4989600
C. 120960
D. None of these

Answer: Option C

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

How many words can be formed with letters of EXAMINATION by taking all the letters at a time in how many of these two vowels do not come together?
Number of ways of arranging these letters =
8! = 10080.
(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = 4! = 12.
2!

How many words can be formed with letters of EXAMINATION by taking all the letters at a time in how many of these two vowels do not come together?
Required number of words = (10080 x 12) = 120960.