Coffee Room

Discuss anything here - everything that you wish to discuss with fellow engineers.

12762 Members

Join this group to post and comment.

# Shortcut to solving probability problems

there are dictionary problems like if the word sachin is to be arranged in a dictionary what is its value. By usual method it can be done but some times it becomes very lengthy , there is a shortcut for this. I forgot that . Can someone please post the shortcut method for it.

*Moved to mathematics section.*

Let us have examples and you mention the method you use to solve them, some maths expert CEans like CP/CT or others can give an easier method then. 👍RVigneshthere are dictionary problems like if the word sachin is to be arranged in a dictionary what is its value. By usual method it can be done but some times it becomes very lengthy , there is a shortcut for this. I forgot that . Can someone please post the shortcut method for it.

Probability appears to be tough but if you practice it is actually easy. If we can have some problem I will definitely try to solve it.

Meanwhile, I am facing similar type of problem. Earlier (2 days back) I knew a method of finding rank of matrix. But thanks to my good memory I forgot!

What to do? Can we arrange some kind of mathematical debate for solving problems?

Meanwhile, I am facing similar type of problem. Earlier (2 days back) I knew a method of finding rank of matrix. But thanks to my good memory I forgot!

What to do? Can we arrange some kind of mathematical debate for solving problems?

for example lets take the word sachin

In dictionary first word is A so remaining words can be arranged in 5! Ways. But we want S , so we will proceed further. Then comes C . Again then remaining words can be arranged in 5! Ways . We will proceed likewise for H I and N

Till now we got 5 times 5! Words . i.e total of 600 words.

So now we get S, as the required word starts from S so we will fix it at first position. Then again we will proceed the same procedure for 2nd alphabet. But fortunately the second required alphabet is A . So we will again fix it. Then comes C but again this is the required word . So proceeding in the same manner we get all the alphabet in the required order.

Hence ultimately we get the required word.

So the rank of word is

5 *5!=600

600+1(the required word)

So if we arrange sachin in dictionary it will be 601st word.

We have a very short method for this . I learnt it when i was in 12th. With that the result can be obtained in 3 or 4 lines. I just forgot that. Please see to that.

In dictionary first word is A so remaining words can be arranged in 5! Ways. But we want S , so we will proceed further. Then comes C . Again then remaining words can be arranged in 5! Ways . We will proceed likewise for H I and N

Till now we got 5 times 5! Words . i.e total of 600 words.

So now we get S, as the required word starts from S so we will fix it at first position. Then again we will proceed the same procedure for 2nd alphabet. But fortunately the second required alphabet is A . So we will again fix it. Then comes C but again this is the required word . So proceeding in the same manner we get all the alphabet in the required order.

Hence ultimately we get the required word.

So the rank of word is

5 *5!=600

600+1(the required word)

So if we arrange sachin in dictionary it will be 601st word.

We have a very short method for this . I learnt it when i was in 12th. With that the result can be obtained in 3 or 4 lines. I just forgot that. Please see to that.

Taking the example :

First arrange them in ascending order: ACHINS

1) Putting in the first letter in the six letter word : S_ _ _ _ _

The number of letters before S in the arrangement is 5.

and no. of blanks in the word got so far is 5 => 5! ways of arranging letters in them

So as of now the ways are 5* 5!

And the arranged letters become: ACHIN (remove the used letters)

2) Putting in the next letter, A => SA _ _ _ _

But there is no letter in front of A So the sum remains unchanged as 5* 5!

And the arranged letters become: CHIN

3)Putting in the next letter, C => SAC _ _ _

But there is no letter in front of C So the sum remains unchanged as 5* 5!

And the arranged letters become: HIN

and so on... there is no addition of values to the sum after that case.

So there are 5*5! words before it

Taking another example :

First arrange them in ascending order: ABKNS

1) Putting in the first letter => B _ _ _ _

Number of letters before B in the arrangement is 1

and number of blanks are 4 => 4! ways

So totally right now the sum is 1 * 4!

And arrangement becomes: AKNS (B removed)

2) Inserting A =>BA _ _ _

But since there is no letter before A in the arrangement the sum does not change.

Arrangement becomes : KNS

3) Inserting N=> BAN_ _

1 letter before N and 2 blanks are there => sum becomes 4! +

Arrangement: KS

They are in the same order and so just add 1

So, the sum becomes 4! + 2! +1

= 27

There are 27 words before it and the rank is 27+1 = 28

Hope it is clear enough. 😀

**SACHIN**.First arrange them in ascending order: ACHINS

1) Putting in the first letter in the six letter word : S_ _ _ _ _

The number of letters before S in the arrangement is 5.

and no. of blanks in the word got so far is 5 => 5! ways of arranging letters in them

So as of now the ways are 5* 5!

And the arranged letters become: ACHIN (remove the used letters)

2) Putting in the next letter, A => SA _ _ _ _

But there is no letter in front of A So the sum remains unchanged as 5* 5!

And the arranged letters become: CHIN

3)Putting in the next letter, C => SAC _ _ _

But there is no letter in front of C So the sum remains unchanged as 5* 5!

And the arranged letters become: HIN

and so on... there is no addition of values to the sum after that case.

So there are 5*5! words before it

Taking another example :

**BANKS**First arrange them in ascending order: ABKNS

1) Putting in the first letter => B _ _ _ _

Number of letters before B in the arrangement is 1

and number of blanks are 4 => 4! ways

So totally right now the sum is 1 * 4!

And arrangement becomes: AKNS (B removed)

2) Inserting A =>BA _ _ _

But since there is no letter before A in the arrangement the sum does not change.

Arrangement becomes : KNS

3) Inserting N=> BAN_ _

1 letter before N and 2 blanks are there => sum becomes 4! +

**1*2!**Arrangement: KS

They are in the same order and so just add 1

So, the sum becomes 4! + 2! +1

= 27

There are 27 words before it and the rank is 27+1 = 28

Hope it is clear enough. 😀

this is an easy one. Thank you very much .😀

But i am afraid this is not the one that i was looking for. But yet this will reduce the work a lot.

Thanks

But i am afraid this is not the one that i was looking for. But yet this will reduce the work a lot.

Thanks

ok... 😀 If you get used to this it is a matter of 30 seconds only. 😀

Will let you know if i find anything else. 😁

Will let you know if i find anything else. 😁

ya definately this is a very good and easy method. Thanks again 😀

why do you add 1 and another 1 in the last part?

I am rather dense.RVigneshthere are dictionary problems like if the word sachin is to be arranged in a dictionary what is its value. By usual method it can be done but some times it becomes very lengthy , there is a shortcut for this. I forgot that . Can someone please post the shortcut method for it.

In my opinion any word of n letters can be represented by a linear matrix of order (1,n). Giving values 1 to 26 for A to Z the matrix can be easily generated.

That for Sachin will be (19,1,3,8,9,14) Sachim will be (19,1,3,8,9,13).

Matrix comparison can be automated in a left to right priority to locate the word in an assemblage of words using a lookup function or some such macro on a database of matrices compiled for each language dictionary.

It is possible that I am completely off base missing some thing in the question

Here are some tips and tricks shared by an IIT Professor, Alok Gupta -

Plus here are some simple formulae that a friend shared with me -

Plus here are some simple formulae that a friend shared with me -