Thanks Chilly for the reply...!
But my question was, if it is not mentioned whether we have to make 'n' lettered words or all possible words, what shall we assume??
I think you should check options. If both the answers are there, then you should pick one in which number of all the possible words are found
1) For ROCKET, if all letters are used, then 6!, else
6 + 6^2 + 6^3 + .. + 6^6 (if repetition is allowed)
Else, 6! + C(6, 5)*5! + C(6, 4)*4! + C(6, 3)*3! + C(6, 2)*2! + 6
2) here TE are fixed, so 6!, if all letters are used, else
Same as in previous case, except that in both the cases their will be addition of 1 when their is no letter between T and E
I think there is no general formula for that, if you have some specific question in mind, then you can ask that.
Q) lets say, we have to distribute 7 distinct balls into 4 similar groups, what are the total ways?
Q.2) we have to distribute 7 similar balls into 4 similar groups, what are the total ways?
12 villages in a district are divided into 3 zones wid 4 villages per zone.The telephone dept. tends to connect the villages with tellephone lines ssuch that every two villages in the same zone are connected wid 3 direct lines and every 2 villages belonging to different zones are connected wid 2 direct lines.How many direct lines are required?????
a.210
b.96
c.54
d.150plzz answer wid explanation
there would be 18 intra-zone lines in each zone= 18*3=54
then there would be 32 inter-zone lines from each zone= 32*3=96.
total lines=54+96=150
plz help me in solving this ques
the no are selected at random ,one at a time ,from two-digit numbers 00,01,02,03,......99 with replacement.an event E occurs only if the product of the two digit of a selected numbers is 18.if four numbers are selected ,find the probability that the event E occurs at least 3 times
a)0.0025
b)0.00025
c)0.025
d)0.25
plz explain me 
Thanks Chilly for the reply...!
But my question was, if it is not mentioned whether we have to make 'n' lettered words or all possible words, what shall we assume??
Thats what I said yaar, there is no way to make out what exactly is asked.
So, you have to solve for both cases and check the options.
If both are in the options then tick the one for all possible words
Q) lets say, we have to distribute 7 distinct balls into 4 similar groups, what are the total ways?
Q.2) we have to distribute 7 similar balls into 4 similar groups, what are the total ways?
First in case of both similar:-
0007 (none of its permutation should be counted again)
0016 (none of its permutations should be counted again)
0025 (no permuatation should be counted again and applicable for all)
0034
0115
0124
0133
0223
1114
1123
1222
So, 11 ways
When objects are different:-
0007 - 1 way
oo16 - C(7, 1) = 7 ways
0025 - C(7, 2) = 21 ways
0034 - C(7, 3) = 35 ways
0115 - C(7, 5) = 21 ways, no need to choose i or the 2 remaining as boxes are identical
0124 - C(7, 4)*C(3, 2)
0133 - C(7, 1)*C(6, 3)/2, as both 3 and 3 are going in identical box
0223 - C(7, 3)*C*4, 2)/2, as both 2 and 2 are going in identical boxes
1114 - C(7, 4)
1123 - C(7, 3)*C(4, 2)
1222 - C(7, 1)*C(6, 2)*C(4, 2)/6, here 6 is in denominator as 2, 2, 2 are going to identical boxes so we have to divided by 3!
Add them to get the total
There is no direct formula for this (generally asked for smaller numbers, so number of cases will be and easy to count)
What is the number of ways of selecting 4 cards, out of a deck of well shuffled playing cards, such that each of the 4 cards has a different face value, and belongs to a different suit?
Eg. if the first card is 2 of hearts, then we cannot get a 2 or a card of hearts again.
What is the number of ways of selecting 4 cards, out of a deck of well shuffled playing cards, such that each of the 4 cards has a different face value, and belongs to a different suit?
Eg. if the first card is 2 of hearts, then we cannot get a 2 or a card of hearts again.
Crabby,
we can choose first card in 52C1 ways.
second in 36C1, third in 20C1 & fourth in 4C1 ways.
total ways = 52C1*36C1*20C1*4c1
Puys, Am I right...!!!
hi...
can someone tell me some shortcut methods to solve permutations and combitations.....its confusing very much i am not able to solve even one problem also.....so pls someone give me some shortcuts to solve it....:-(:-(:-(
Crabby,
we can choose first card in 52C1 ways.
second in 36C1, third in 20C1 & fourth in 4C1 ways.
total ways = 52C1*36C1*20C1*4c1
Puys, Am I right...!!!
Can you please explain your approach?
I went like this: 52 for the first card,
36 for the second,
for the third,
there are 2 suits available but only 11 face values, so 22 ways to choose,
and for the fourth, there is only 1 suit, and 10 ways to pick a face value that has not yet been picked.
52.36.22.10
However, the OA in Arun Sharma's book is given as 13.12.11.10:wow:
please answer this and explain the difference
Q. in how many ways can 12 different books be divided equally among
(i)4 persons
(ii)4 sets or group
please answer this and explain the difference
Q. in how many ways can 12 different books be divided equally among
(i)4 persons
(ii)4 sets or group
It can be done like this.
Arrange 12 books and divide in 4 parts.
(i) 12 books can be arranged in 12! ways.
And divided in 4 parts. But in each part there can be 3! arrangements which will be overcounting. So, we need to divide by 3! 4 times.
So, total ways for part (i) = 12! / (3!)^4
(ii) Now, in case of sets, we need to further divide by 4!. As arragement to 4 persons will be 4! times the arrangement in sets.
So, answer to part (ii) will be 12! /
In how many ways a cube can be painted with three colors red,gree and blue??
Ans is 30
It can be done like this.
Arrange 12 books and divide in 4 parts.
(i) 12 books can be arranged in 12! ways.
And divided in 4 parts. But in each part there can be 3! arrangements which will be overcounting. So, we need to divide by 3! 4 times.
So, total ways for part (i) = 12! / (3!)^4
(ii) Now, in case of sets, we need to further divide by 4!. As arragement to 4 persons will be 4! times the arrangement in sets.
So, answer to part (ii) will be 12! /
Thanks for the explanation but i didn't understand the difference between sets and person....i.e. why, in the case of sets it has been divided by 4!
nam2aspire SaysThanks for the explanation but i didn't understand the difference between sets and person....i.e. why, in the case of sets it has been divided by 4!
Take a simpler example.. You have 2 books.. A and B.
If you want to assign one book to each person say P1 and P2.
So, we have two ways:
P1 gets book A and P2 gets book B.
Or P1 gets book B and P2 gets book A.
Now ifneed to form two sets of books.
Only 1 way.. A and B. Here, (A and B) or (B and A) are same.
So, one way to form sets.
So, to form sets, we need to divide by 2!.
So, in above example, we need to divide by 4!.
Can you please explain your approach?
I went like this: 52 for the first card,
36 for the second,
for the third,
there are 2 suits available but only 11 face values, so 22 ways to choose,
and for the fourth, there is only 1 suit, and 10 ways to pick a face value that has not yet been picked.
52.36.22.10
However, the OA in Arun Sharma's book is given as 13.12.11.10:wow:
In This approach you have also arranged those cards among themselves but this is not required in the answer .If u see both the answers carefully
52.36.22.10=4!(13.12.11.10)
So when u take first card among 52 it means your open to arrange those cards among themselves .But when u selected the first card from 13 then ur not arranging among themselves i hope its clear
Some one please explain me the concept :
'How to arrange 5 beads on a necklace ?'
Can you please explain your approach?
I went like this: 52 for the first card,
36 for the second,
for the third,
there are 2 suits available but only 11 face values, so 22 ways to choose,
and for the fourth, there is only 1 suit, and 10 ways to pick a face value that has not yet been picked.
52.36.22.10
However, the OA in Arun Sharma's book is given as 13.12.11.10:wow:
ohh.. u r rt... my mistake ...!!
Total ways= 52.36.22.10
Anyways, it doesn't match the solution in the book.
Can anyone explain the solution in Arun Sharma..!!!
Some one please explain me the concept :
'How to arrange 5 beads on a necklace ?'
it should be (5-1)!/2
The logic applied is same as is for circular permutation with only diff' that bcoz of hook in the necklace, you get 2 arrangements, clockwise & anticlockwise, hence divided by 2.
There are 12 cities in a state . Each of them is to be connected with 3 other cities using cable wire . In how many ways this can be accomplished ??
There are 12 cities in a state . Each of them is to be connected with 3 other cities using cable wire . In how many ways this can be accomplished ??
Each city has 3 wires associated with it......so 12*3=36......
But each wire is counted twice......hence there are 18 wires......
Now for each wire you need to find a pair of cities......
Total number of possible pairs=12C2=66......
18 pairs can be chosen in 66C18 ways......