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, . 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!.
but here each person is getting 3 books and each set would also have suppose the same three books like that of the four persons P1 P2 P3 AND P4 have (let ABC DEF GHI JKL).Hence why in the case of sets it would appear like this concept as we are not to find its PERMUTATION!!!!!!!??????
hey guys could someone help me in solving the following two problem with clear explanation.
a) In how many ways can a selection be made of 5 letters out of 5 As,4 Bs,3 Cs,2 Ds and 1 E?
b)From 4 gentlemen and 4 ladies a committee of 5 is to be formed with atleast 3 women such that atleast one woman holds the post of either a president or a vice president
How many 4 letter words can b formed using letters A,B,C,D,E,F and G such that at least one letter of the word is a vowel?
I followed the following method- There are 2 vowels and 5 consonants.Since I have to choose atleast one vowel so 2C1 .Now after the selection I have 6 alphabets left(1 vowel & 5 consonants) so I choose 3 from them in 6C3 ways.And all this can be arranged in 4! ways.
So answer comes to be 2c1*6c3*4! = 960
But, the answer is 720.I also got 720 using other approach. Plz tell me what is the flaw in the above logic.
How many 4 letter words can b formed using letters A,B,C,D,E,F and G such that at least one letter of the word is a vowel?
I followed the following method- There are 2 vowels and 5 consonants.Since I have to choose atleast one vowel so 2C1 .Now after the selection I have 6 alphabets left(1 vowel & 5 consonants) so I choose 3 from them in 6C3 ways.And all this can be arranged in 4! ways.
So answer comes to be 2c1*6c3*4! = 960
But, the answer is 720.I also got 720 using other approach. Plz tell me what is the flaw in the above logic.
Q1.5 letter word to be constructed from MATHEMATICS,in how many words will E occur?
Ans:6990
Q2.30 identical prizes can be distributed among 4 boys such dat each boy get an odd number of preizes and a particular boy got 3k prizes where "k" is a natural number?
Q3.10 persons seated in a row find d ways in which 4 persons can be selected such dat no 2 of them are sitting nex to each oder?
Ans:35
Q4.Total number of ways 4 couples can be seated around circular table such that men and women are seated alternately and no man seated beside his wife?
but here each person is getting 3 books and each set would also have suppose the same three books like that of the four persons P1 P2 P3 AND P4 have (let ABC DEF GHI JKL).Hence why in the case of sets it would appear like this concept as we are not to find its PERMUTATION!!!!!!!??????
That is because, we form two sets of one book each; A and B. So, there is only 1 way to form sets. However, these two sets can be given to 2 people in 2! ways.
So, basically, number of ways books sets can be given = number of ways sets can be formed * n! .... where n is number of persons or number of sets.
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......
I have a question in bold part.. Each city is connected only to 3 other cities.. How do we make this sure? Say, in 66 pairs, there will be 11 pairs for city A. In 66C18, those all 11 pluas 7 others can be selected. So, I think, it will violate the condition that each city is connected only 3 other cities.
hello puys! i have this permutation and combination problem which has been answered in this pagalguy thread ALREADY. BUT i noticed that the answer provided in the forum DID NOT match with the solutions provided with the question. please help me with the approach.
How many 5 digit numbers exist having exactly two 4s in them?
PS- also, from what i observed, answers come out to be different depending upon whether the "fixed" digit chosen for permuting is...either "unit place" or "ten thousands" place ??:shocked:
hello puys! i have this permutation and combination problem which has been answered in this pagalguy thread ALREADY. BUT i noticed that the answer provided in the forum DID NOT match with the solutions provided with the question. please help me with the approach.
How many 5 digit numbers exist having exactly two 4s in them?
PS- also, from what i observed, answers come out to be different depending upon whether the "fixed" digit chosen for permuting is...either "unit place" or "ten thousands" place ??:shocked:
hey! hii! okay! as per the answer i have, it should be 9400.. and not 1344.
though i would be keen to know the logic that you followed?
The anwer should be 9400.Here is how . First, divide the case into two cases 1 that starts with 4 i.e. 4 is the most significant digit.and case 2 that does'nt case 1 -4C1 *10*10*10 =4000 Case2- 4C2 *9*10*10=5400 Total=9400
The anwer should be 9400.Here is how . First, divide the case into two cases 1 that starts with 4 i.e. 4 is the most significant digit.and case 2 that does'nt case 1 -4C1 *10*10*10 =4000 Case2- 4C2 *9*10*10=5400 Total=9400
yeah, the answer is correct. but i am doubtful about the approach. especially because the answers are different when (a) 4 is considered to start at units digit OR (b) 4 is considered to start at ten thousands digit. Also, I think the ten thousands digit will have "9" options and not "10". reason : "0" cant be at ten thousands place for a 5 digits number.
Q1.5 letter word to be constructed from MATHEMATICS,in how many words will E occur?
Ans:6990
Q2.30 identical prizes can be distributed among 4 boys such dat each boy get an odd number of preizes and a particular boy got 3k prizes where "k" is a natural number?
Q3.10 persons seated in a row find d ways in which 4 persons can be selected such dat no 2 of them are sitting nex to each oder?
Ans:35
Q4.Total number of ways 4 couples can be seated around circular table such that men and women are seated alternately and no man seated beside his wife?
The anwer should be 9400.Here is how . First, divide the case into two cases 1 that starts with 4 i.e. 4 is the most significant digit.and case 2 that does'nt case 1 -4C1 *10*10*10 =4000 Case2- 4C2 *9*10*10=5400 Total=9400
Could you please explain the bold part? as per question there should be exactly two 4s.so in case 1 i think it should have been =(4*9*9*9) I am not able to grasp this concept...please make it clear.
Could you please explain the bold part? as per question there should be exactly two 4s.so in case 1 i think it should have been =(4*9*9*9) I am not able to grasp this concept...please make it clear.
The anwer should be 9400.Here is how . First, divide the case into two cases 1 that starts with 4 i.e. 4 is the most significant digit.and case 2 that does'nt case 1 -4C1 *10*10*10 =4000 Case2- 4C2 *9*10*10=5400 Total=9400
hey! hii! okay! as per the answer i have, it should be 9400.. and not 1344.
though i would be keen to know the logic that you followed?
Hello! Puys! i just found out that the approach followed in the solutions provided with this question is WRONG. the answer "9400" has been calculated for "at least two 4s" (which is not true according to the question). ideally its should be (for EXACTLY two 4s)..
Hello! Puys! i just found out that the approach followed in the solutions provided with this question is WRONG. the answer "9400" has been calculated for "at least two 4s" (which is not true according to the question). ideally its should be (for EXACTLY two 4s)..
{4C1*9*9*9} + {4C2*8*9*9}
Yeah!! For Exactly two 4s, it should be Case 1- when 4 is placed at most significant digit 4C1*9*9*9 (any one of the four places in 4C1 ways and the remaining three places with 9 ways as 4 cant be placed more than twice) Case 2- when 4 is not placed at most significant digit 8*4C2*9*9 (1st place can be filled with any 8 other than 4 & 0. two places for 4 can be selected in 4C2 ways and remaining two places in 9 ways) ANS: 6804
