Q: The number of integral solutions for the equation a+b+c+d = 12 , where ( a,b,c,d ) >= -1 is :
a) 19C3
b) 18C4
c) 20C4
d) 16C2
e) 17C1
Algebra
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Let us assume new variables p, q, r, and s such that, p = (a + 1), q = (b + 1), r = (c + 1) and s = (d + 1). (Why I did it? See note.)Aman verma wrote:Q: The number of integral solutions for the equation a+b+c+d = 12 , where ( a,b,c,d ) >= -1 is :
Hence, p, q, r, s ≥ 0
And, (a + b + c + d) = 12 ---> (p + q + r + s) = 16
So, we need to find out the number of integral solutions for the equation (p + q + r + s) = 16, where p, q, r, s ≥ 0
Now, the problem can be rephrased as "In how many ways sixteen 1s and three +s can be arranged in a row?"
Number of way to do so = (16 + 3)!/[(16!)*(3!)] = 19!/[(19 - 3)!*3!] = 19C3
The correct answer is A.
Note #1: I changed the variables to make them countable items. It is difficult to interpret negative values as countable items.
Note #2: In general the formula to determine number of non-negative integral solutions of the equation (a + b + c + d + ...) = n is (n + r - 1)C(r - 1), where r is the number of variables. Here, n = 16 and r = 4.
Note #3: This method is popularly known as the "inserting stick" or "separator" method which is very useful for a large number of complex combinatorics problems. For more use cases of this method, read the following posts made by Rahul,
https://www.beatthegmat.com/12-t70715.html#319654
https://www.beatthegmat.com/12-t70715.html#319982
https://www.beatthegmat.com/oranges-t70953.html#320828
Anju Agarwal
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Hi Anju,
Thanks for the response. That link to the Rahul's thread was very helpful !
Thanks for the response. That link to the Rahul's thread was very helpful !
800. Arjun's-Bird-Eye
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If you're interested, here's another practice question along the same lines: https://www.beatthegmat.com/very-tricky- ... 25349.html
Cheers,
Brent
Cheers,
Brent
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Since abcd ≥ -1:Aman verma wrote:Q: The number of integral solutions for the equation a+b+c+d = 12 , where ( a,b,c,d ) >= -1 is :
a) 19C3
b) 18C4
c) 20C4
d) 16C2
e) 17C1
a≥-1, b≥-1, c≥-1, and d≥-1.
Thus:
a+1≥0, b+1≥0, c+1≥0, and d+1≥0.
Since a+b+c+d = 12, the following must also be true:
(a + 1) + (b + 1) + (c + 1) + (d + 1) = 16.
Here, a sum of 16 must be DISTRIBUTED among 4 NONNEGATIVE integers: a+1, b+1, c+1, and d+1.
This question is no different from the following:
How many ways can 16 identical marbles be distributed among 4 children a+1, b+1, c+1, and d+1?
Use the SEPARATOR method.
16 identical marbles are to be separated into -- at most -- 4 groupings.
Thus, we need 16 marbles and 3 separators:
OOO|OO|OOOOO|OOOOOO
Each arrangement of the elements above represents one way to distribute the 16 marbles among 4 children a+1, b+1, c+1, and d+1:
OOOO|OOO|OOO|OOOOOO = a+1 gets 4 marbles, b+1 gets 3 marbles, c+1 gets 3 marbles, d+1 gets 6 marbles.
OOOOOOOOO|||OOOOOOO = a+1 gets 9 marbles, d+1 gets 7 marbles.
OOOOOOOOOOOOOOOO||| = a+1 gets all 16 marbles.
And so on.
To count all of the possible distributions, we simply need to count the number of ways to arrange the 19 elements above (the 16 identical marbles and the 3 identical separators).
The number of ways to arrange 19 elements = 19!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 16! (the number of ways to arrange the 16 identical marbles) and by 3! (the number of ways to arrange the 3 identical separators):
19!/(16! 3!) = (19*18*17)/3! = 19C3.
The correct answer is A
Here's a similar problem:
An odd value can be represented as 2x + 1.Find the total number of ways in which 35 identical marbles can be distributed among 5 boys such that each boy gets odd number of marbles?
1) 39C4
2) 19C4
3) 15C4
4) 20C4
5) 25C4
Since each of the 5 boys receives an ODD number of marbles, and the total number of marbles is 35, we can represent the sum as follows:
(2a + 1) + (2b + 1) + (2c + 1) + (2d + 1) + (2e + 1) = 35.
2(a+b+c+d+e) + 5 = 35
a+b+c+d+e = 15.
Now the question becomes:
How many ways can 15 identical marbles be distributed among 5 children a, b, c, d and e?
Here, we need 15 marbles and 4 separators:
OOO|OO|OOOOO|OO|OOO
Each arrangement of the elements above represents one way to distribute the 15 marbles among 5 children a, b, c, d and e:
OOOO|OOO|OOO|OO|OO = a gets 4 marbles, b gets 3 marbles, c gets 3 marbles, d gets 2 marbles, e gets 2 marbles.
OOOOOOOO||||OOOOOOO = a gets 8 marbles, e gets 7 marbles.
OOOOOOOOOOOOOOO|||| = a gets all 15 marbles.
And so on.
To count all of the possible distributions, we simply need to count the number of ways to arrange the 19 elements above (the 15 identical marbles and the 4 identical separators).
The number of ways to arrange 19 elements = 19!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 15! (the number of ways to arrange the 15 identical marbles) and by 4! (the number of ways to arrange the 4 identical separators):
19!/(15! 4!) = = (19*18*17*16)/4! = 19C4.
The correct answer is B.
It should be noted that both of these problems are WAY beyond the scope of the GMAT.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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