permutations and combinations

Firstly,

Here, out of a total of 7 letters, 5 can be selected for empty slots in 7C5 ways.

You have already taken two letters a and b out of the set. So you don't have to choose 5 out of 7, Instead 5 out of 5.

I lost my pinkies!! Nice Khyati!

it has really got targetfighty nowadays...... don"t respond to this ...only a joke:)

The solution that I gave in my post #13 is for the case when we have to take both permutation and combination together I think.

Yippie!!!

Thanks a lot Sir. [1]

I din said there is something wrong with your post, but you are not taking into account other possible cases.

You said in your post that the cases where b come before than a is neglected, but I have toh included them in my calculations, seems u din read my posts carefully.

@ Swordfish,

1680 is the correct answer.

When it is clearly mention in the question that we are dealing here with the PERMUTATIONS, then why are you using combination concept here. ?

Secondly, you cannot use letters a and b while making the words because it will result in the repetition of the letters in the words.

Hope this makes the things clear. [1]

I used combinations and then arranged them in so and so ways. So it is nothing but permutation.

#32 is same as selecting and then arranging 6 out of {a,a,a,b,b,b,c,c,c}

³C₂6!3!3! + ³C₁ ²C₁6!3!2! + 6!2!2!2! =510 ?(mostly it is wrng :P )

1. if only 7 lettered words are needed ,

a x y z b → ⁵C₃ 3! 2! 3! = 720

⁵C₃ for selecting 5 letters from c,d,e,f,g to put betweeb a and b , 3! to arrange these 3 selected , 2! to arrange a and b , 3 ! again to arrange the whole word assuming (axyz) as one letter

similarly
a x y b → ⁵C₂ 2!2!4! = 960

hence total = 1680

3) how many 15 letter arrangements of 5A`s,5B`s,5C`s have no A's in the first five letters, no B's in the next 5 letters and no C's in the last five letters.

And from the question, what I interpret is that we are told to from a letter consisting of all digits (and not 4 or 5) but in such a way that either 2 or 3 digits come in between a and b.

Eg.

Case 1: Fix two places between a and b, then we can permute the position of this 'set' as well the the letter between it.

a _ _ b _ _ _ ; _ a _ _ b _ _ ; _ _ a _ _ b _ ; _ _ _ a _ _ b

Apply permutations for blank ones and then probably *4

Case II: Fix three places between a and b, then we can permute the position of this 'set' as well the the letter between it.

a _ _ _ b _ _ ; _ a _ _ _ b _ ; _ _ a _ _ _ b

Apply permutations for blank ones and then probably *3

Answer should be Case I + Case II

NOTE:

b _ _ a is also very much possible :)

This case is being neglected here.

If it was also the case then we would have in Case I (*8 inplace for *4) and (*6 in place for *3)

Well, We may have two cases :

Case 1. Fix two places between a and b, then we can permute the position of this 'set' as well the the letter between it.
Case 2. Fix three places between a and b, then we can permute the position of this 'set' as well the the letter between it.

Therefore, the total no. should be Case 1+Case 2.

Am I correct? Please tell me?

well thats correct in a crude way but u must be going like this:

what lettered word are u talking about

from question its clear that , minimum has to be a 4 lettered work

so u have 4,5,6,7 letered words possible

now for a 4 lettered

a_ _ b

for 5 lettered, u have the above set as well as one more blank space avalable so it can be

_a_ _ b or a_ _ b _

hope u can proceed on these lines now

i forgot a point in my above post

b _ _ a is also very much possible :)

so include it too :)

I think I have said the same too. But Since I don't know how to compute Permutations and Combinations. :)

For this question there can be 4 cases which are as follows

1)when there are 4 letter words,

a _ _ b

from among 5 remaining letters 2 can be taken in 5P2 ways which can be arranged themselves in 2! ways and a and b can also be arranged among themselves in 2 ways, so

5P2*2!*2! = 80

2)5 letter words, here can be 3 cases too which are as follow:-

A) a _ _ b _
B) a _ _ _ b
c) _ a _ _ b

letters can be arranged here as

(5P3*3!*2)*3 = 2160.

3) when 6 lettered words are formed

A) a _ _ _ _ b
B) a _ _ _ b _
C)a _ _ b _ _
D)_ a _ _ b _
E)_ a _ _ _ b
F) _ _ a _ _ b

here the letters can be arranged as

(5P4*4!*2)*6 = 34560

4)when 7 lettered word is formed

A) a _ _ _ _ _ b
B) a_ _ _ _ b _
C) a _ _ _ b _ _
D) a _ _ b _ _ _
E)_ a _ _ _ _ b
F) _ _ a _ _ _ b
G) _ _ _ a _ _ b

(5P5*5!*2) *7 =201600

So the total no. of permutations are = 80 + 2160 +34560+ 201600 = 238400

either two or three letters between a and b?

Did you pay emphasis to this line of the question. Your 6th and 7th cases don't show such.

That's why I had only two cases (2 or 3 letters between a and b)

I request Nishant sir not to solve this one, let this be for us users.

Well Vivek I have written 2 inside bracket which is for the arrangement of a and b, so there is no need of the correction that you have told.