71Firstly,
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!
1)
How many permutations of the letters a, b, c, d, e, f, g have either two or three letters between a and b?
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36 Answers
7171You 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.
Read the line after that!! oO
1it has really got targetfighty nowadays...... don"t respond to this ...only a joke:)
11Yeah I agree I was wrong, there would only be 7 lettered words since we are dealing with permutations here.
Hope this one should be the correct answer. It is as follows.
there would be only one case,i.e,7 lettered words
a) a _ _ _ b _ _
b) a _ _ b _ _ _
c) _ _ a _ _ _ b
d) _ _ _ a _ _ b
e) _ a _ _ b _ _
f) _ a _ _ _ b _
g) _ _ a _ _ b _
5! * 7 * 2
Here the remaining letters c, d, e, f, g can be arranged in 5!,i.e, 120 ways which would be multiplied by 7 as there are 7 above mentioned cases and 2 are the ways in which a and b can be arranged among themselves.
So the answer comes out to be,
5!*7*2 = 1680.
11The solution that I gave in my post #13 is for the case when we have to take both permutation and combination together I think.
1I would like to add my answer-
1)2 letters between a and b
a_ _ b _ _ _
Here, out of a total of 7 letters, 5 can be selected for empty slots in 7C5 ways.
Now these can be arranged themsleves in 4! x 2! x 2! ways (treating a_ _b as one group and a,b can arrange themselves in 2 ways and 2 slots in 2 ways)
So 7C5 x 4! x 2! x 2! = 2016 ( 2! for a and b )
2)3 letters between a and b
a_ _ _ b _ _
Again, out of 7 letters, 5 can be selected in 7C5 ways.
Treating a _ _ _ b as one group, these can be arranged in 3! ways with a and b arranging themselves in 2! ways and 3 slots in 3! ways.
So 7C5 x 3! x 3! x 2! = 1512
Total = 3528
11I 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.
11@ 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]
392) A man has 3 friends. In how many ways can he invite one friend for dinner every night for 6 consecutive nights so that no friend is invited more than 3 times.
1I used combinations and then arranged them in so and so ways. So it is nothing but permutation.
23#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 )
231. 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
393) 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.
71And 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)
71Well, 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?
39well 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
39i forgot a point in my above post
b _ _ a is also very much possible :)
so include it too :)
71I think I have said the same too. But Since I don't know how to compute Permutations and Combinations. :)
11answer wrong hai mera, usme bahut kuchh baki hai.I considered 2 case ek saath woh bhi hal half
11For 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
71either 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)
11I request Nishant sir not to solve this one, let this be for us users.
11Well 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.
11Yeah I made the mistake and now am correcting it.
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
here the letters can be arranged as
(5P4*4!*2)*5 = 28800
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 now am getting the answer as 232640