Hi! I am a Computer Science Engineer studying math for an exam

Hi, i am a Computer Science Engineer.. I'm studying math for an exam and never been good at it.. now trying my best bottom up!! need help in a lot of problems.. the first one is here :), thanx a lot!

This is a question from the ISC book M. L. Agarwal. Please tell this approach to this question.

How many words can be made with letters of the word INTERMEDIATE if

(i) the words neither begin with I nor end with E

(ii) the relative order of vowels and consonants does not change.

Ans. (i) 83X((10!)/24) (ii) 21600

Re: Hi! I am a Computer Science Engineer studying math for an exam

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Originally Posted by

**rachitmagon** Hi, i am a Computer Science Engineer.. I'm studying math for an exam and never been good at it.. now trying my best bottom up!! need help in a lot of problems.. the first one is here :), thanx a lot!

This is a question from the ISC book M. L. Agarwal. Please tell this approach to this question.

How many words can be made with letters of the word INTERMEDIATE if

(i) the words neither begin with I nor end with E

(ii) the relative order of vowels and consonants does not change.

Ans. (i) 83X((10!)/24) (ii) 21600

What are you asking?

Re: Hi! I am a Computer Science Engineer studying math for an exam

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Originally Posted by

**Plato** What are you asking?

The approach to deal with this question, i know the numeric answer from a source but i don't know how to go about it.

Re: Hi! I am a Computer Science Engineer studying math for an exam

Quote:

Originally Posted by

**rachitmagon** The approach to deal with this question, i know the numeric answer from a source but i don't know how to go about it.

Quote:

Originally Posted by

**rachitmagon** How many words can be made with letters of the word INTERMEDIATE if

(i) the words neither begin with I nor end with E

Ans. (i) 83X((10!)/24)

Because of repeated letters the total number of arrangements is $\displaystyle \frac{12!}{(2!)^2(3!)}$.

Now the answer the question you need to find the number of those that begin with *I* plus the number that end with *E* minus the number beginning with *I* **and** end with *E*.

So $\displaystyle \frac{11!}{(2!)(3!)}+\frac{11!}{(2!)^3}-\frac{10!}{(2!)(2!)}$.

Because this is not a free tutorial site, that is all you get.

Re: Hi! I am a Computer Science Engineer studying math for an exam

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Originally Posted by

**Plato** Because this is not a free tutorial site, that is all you get.

That helps a lot!! Thank you! I think I understand the second part also now

Answer for the second part: since they do not change their relative positions we only change the vowels and consonants with each other (6 each)

$\displaystyle \frac{6!}{(2!)(3!)}*\frac{6!}{(2!)}$