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23 Feb 2021, 03:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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There are (p + q) different books on different topics in Mathematics (p ≠ q)
If:
L = The number of ways in which these books are distributed between two students X and Y such that X get p books and Y gets q books.
M = The number of ways in which these books are distributed between two students X and Y such that one of them gets p books and another gets q books.
N = The number of ways in which these books are divided into two groups of p books and q books.
Then, which of the following must be true?
(A) L = M = N
(B) L = 2M = 2N
(C) 2L = M = 2N
(D) L = M = 2N
(D) L = M = 3N
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
If:
L = The number of ways in which these books are distributed between two students X and Y such that X get p books and Y gets q books.
M = The number of ways in which these books are distributed between two students X and Y such that one of them gets p books and another gets q books.
N = The number of ways in which these books are divided into two groups of p books and q books.
Then, which of the following must be true?
(A) L = M = N
(B) L = 2M = 2N
(C) 2L = M = 2N
(D) L = M = 2N
(D) L = M = 3N
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
23 Feb 2021, 04:14
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Bunuel
L = The number of ways in which these books are distributed between two students X and Y such that X get p books and Y gets q books.
This is same as N, where we have to divide p+q books into p books and q books.
We have to find the combinations of p books out of p+q books, the other is already created..
=\(^{(p+q)}C_p*^{(p+q-p)}C_q\)
=\(^{(p+q)}C_p*^{(q)}C_q\)
=\(^{(p+q)}C_p*1\)
=\(\frac{(p+q)!}{p!(p+q-p)!}\)...
=L=N
Now, M = The number of ways in which these books are distributed between two students X and Y such that one of them gets p books and another gets q books.
Let X get p books and Y get q books
= \(^{(p+q)}C_p*^{(p+q-p)}C_q\)
=\(^{(p+q)}C_p*^{(q)}C_q\)
=\(^{(p+q)}C_p*1\)
=\(\frac{(p+q)!}{p!(p+q-p)!}=\frac{(p+q)!}{p!(q)!}\)
Now, let X get q books and Y get p books
= \(^{(p+q)}C_q*^{(p+q-q)}C_p\)
=\(^{(p+q)}C_q*^{p}C_p\)
=\(^{(p+q)}C_q*1\)
=\(\frac{(p+q)!}{q!(p)!}\)
Add the two = \(\frac{(p+q)!}{p!(q)!}+\frac{(p+q)!}{q!(p)!}=2*\frac{(p+q)!}{p!(q)!}=2L=2N\)
Thus 2L=M=2N
C
08 Sep 2021, 07:57
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Actually not TOO bad if you assume some easy to numbers and use the concept of Distributing Different Items to Different Groups and the concept of Distributing Different Items to Identical Groups (i.e., placing the books into "Identical Stacks")
The Algebraic way is helpful to learn the underlying strategy to answer the question and the concept. However, writing that many Letters does not seem appealing at the moment : - )
Let there be 5 Total Books:
P can NOT equal Q so Let:
P = 2 books
Q = 3 Books
all books are different, so it DOES MATTER WHICH book goes to which person (X or Y) or which book ends up in which indistinguishable "stack"
*note* "Indistinguishable" means the groups are NOT Unique Labeled -----> Breaking up the books so 3 go into the stack on the Left and 2 go into the stack on the Right----- is the SAME as ----- if you were to distribute the books such that 2 go into the stack on the Left and 3 go into the stack on the Right.
(1st) L
we are distributing the books so that X will get P = 2 of the unique books and Y will get Q = 3 of the unique books
step 1: how many different ways can we select 2 out of the 5 books and then select 3 out of the remaining so that we have TWO "indistinguishable" stacks of 2 books and 3 books
(5 books) Choose (2) = (5 c 2)
and
ofr remaining 3 books, Choose 3 = (5 c 3)
same as just choosing the 2 books because when you make selections of different groups of 2 books out of 5, the remaining 3 books are automatically grouped
5 c 2 = 5! / 2!3! = 5*4 / 2 = 10 WAYS
now, the Key here is that since X will get the P (2) books and Y will get the Q (3) books ----> we do NOT have to shuffle around/arrange these indistinguishable stacks of 2 books and 3 books to X and Y
Total Unique Distributions = 10 = L
L = 10
(2nd) M = no. of ways the P and Q books are distributed among X and Y so one student gets the P books and the other student gets the Q books
the Logic is exactly the same as above EXCEPT NOW, it matters which indistinguishable stack goes to X and which goes to Y.
so for every 1 of the 10 unique distributions above:
case 1: X could receive the P # of books and Y could receive the Q # of books
OR
case 2: X could receive the Q # of books and Y could receive the P # of books
thus, the number of unique distributions is: (5 c 2) * 2! = 20 Ways
M = 20
(3rd) N = the no. of ways we can divide the books into "indistinguishable" stacks of P and Q books - this is dividing the books into groups that are NOT uniquely labeled (the only difference being the AMOUNT of different books that go into these stacks)
Unlike in Scenario (2nd) above in which we had to arrange/shuffle around the stacks to 2 different people, here we do NOT have to do any such thing
we just need to figure out how many different ways can we divide the books so that P (2 books) and Q (3 books) are separated into these two indistinguishable stacks.
the answer will be the same as scenario (1st)
(5 c 2) * (3 c 3) = 10 ways
N = 10
SUMMARY:
L = 10
M = 20
N = 10
thus: M is TWICE of L and M is TWICE of N
M = 2L
M = 2N
and L and N are EQUAL
L = N-----> which means that: 2L = 2L
and setting up the equation:
2L = M = 2N
Answer Choice (C)
The Algebraic way is helpful to learn the underlying strategy to answer the question and the concept. However, writing that many Letters does not seem appealing at the moment : - )
Let there be 5 Total Books:
P can NOT equal Q so Let:
P = 2 books
Q = 3 Books
all books are different, so it DOES MATTER WHICH book goes to which person (X or Y) or which book ends up in which indistinguishable "stack"
*note* "Indistinguishable" means the groups are NOT Unique Labeled -----> Breaking up the books so 3 go into the stack on the Left and 2 go into the stack on the Right----- is the SAME as ----- if you were to distribute the books such that 2 go into the stack on the Left and 3 go into the stack on the Right.
(1st) L
we are distributing the books so that X will get P = 2 of the unique books and Y will get Q = 3 of the unique books
step 1: how many different ways can we select 2 out of the 5 books and then select 3 out of the remaining so that we have TWO "indistinguishable" stacks of 2 books and 3 books
(5 books) Choose (2) = (5 c 2)
and
ofr remaining 3 books, Choose 3 = (5 c 3)
same as just choosing the 2 books because when you make selections of different groups of 2 books out of 5, the remaining 3 books are automatically grouped
5 c 2 = 5! / 2!3! = 5*4 / 2 = 10 WAYS
now, the Key here is that since X will get the P (2) books and Y will get the Q (3) books ----> we do NOT have to shuffle around/arrange these indistinguishable stacks of 2 books and 3 books to X and Y
Total Unique Distributions = 10 = L
L = 10
(2nd) M = no. of ways the P and Q books are distributed among X and Y so one student gets the P books and the other student gets the Q books
the Logic is exactly the same as above EXCEPT NOW, it matters which indistinguishable stack goes to X and which goes to Y.
so for every 1 of the 10 unique distributions above:
case 1: X could receive the P # of books and Y could receive the Q # of books
OR
case 2: X could receive the Q # of books and Y could receive the P # of books
thus, the number of unique distributions is: (5 c 2) * 2! = 20 Ways
M = 20
(3rd) N = the no. of ways we can divide the books into "indistinguishable" stacks of P and Q books - this is dividing the books into groups that are NOT uniquely labeled (the only difference being the AMOUNT of different books that go into these stacks)
Unlike in Scenario (2nd) above in which we had to arrange/shuffle around the stacks to 2 different people, here we do NOT have to do any such thing
we just need to figure out how many different ways can we divide the books so that P (2 books) and Q (3 books) are separated into these two indistinguishable stacks.
the answer will be the same as scenario (1st)
(5 c 2) * (3 c 3) = 10 ways
N = 10
SUMMARY:
L = 10
M = 20
N = 10
thus: M is TWICE of L and M is TWICE of N
M = 2L
M = 2N
and L and N are EQUAL
L = N-----> which means that: 2L = 2L
and setting up the equation:
2L = M = 2N
Answer Choice (C)
Bunuel
