Sums Of 10
How many ways are there of adding together two or more integers from
1
to
9
to make
1
0
? For example,
1
+
9
=
1
0
,
9
+
1
=
1
0
,
2
+
8
=
1
0
,
8
+
2
=
1
0
, and
2
+
1
+
2
+
1
+
1
+
2
+
1
=
1
0
. (Sums that contain the same numbers but in a different order are considered to be different.)
The answer is 511.
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3 solutions
I solved it the exact same way! :D Good solution!
could u xpalin the solution by sum other example...coz this example i culd nt understand.....thnx
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Do you understand the ball and stick representation? For example a number is just represented by the number of balls between 2 sticks or on the left of the stick (if it is the first stick - reading from left to right). For example, we represent 6 as 0000001 because there as 6 balls to the left of the stick. Another example will be 000100001. Here 2 numbers are being represented, first is 3 (3 balls to left of first stick) and 4 (since 4 balls between the 2 sticks). Also, note that the remaining balls after the last stick will be the last number (to be added). E.g. 0001000001000. The 3 numbers represented are 3, 5 and 3. Can u see why?
So, in this question, we make use of this simplification to want to find out the number of ways to add some numbers (integers from 1 to 9) to get 10. This is just equivalent to finding the number of ways to divide 1 0 balls into a certain number of parts. So, we have 1 0 balls represented by 1 0 zeros. To illustrate what I mean before we continue, here's another example. 0000001010100 represents the numbers (once again from left to right) 6 , 1 , 1 , 2 . Therefore, this in turns represent a way to add to the sum 1 0 = 6 + 1 + 1 + 2 .
Then notice that we have 2 choices to place a stick (or 1) in between each 0 since we let there be 1 0 balls at first. There are 9 such gaps between 2 consecutive balls. So, there are in total 2 9 ways to place sticks between the balls (or to add numbers). However, here we minus 1 because we cannot not put any sticks (or put 0 sticks). Therefore, the answer is 2 9 − 1 = 5 1 1 .
Is it clearer now?
I think, you missed it by 5+5 method. I got 510, confused.
yeah correct answer
great
Wow, thanks for the fantastic solution.
C a s e 1 : a + b = 1 0 , a > 0 , b > 0 C a s e 2 : a + b + c = 1 0 , a > 0 , b > 0 , c > 0 . . . T i l l C a s e 9 N u m b e r O f S o l u t i o n s F o r x 1 + x 2 + x 3 + . . . . . . . . . + x n = a , x k > 0 , 1 ≤ k ≤ n = a − 1 C n − 1 N u m b e r O f S o l u t i o n s F o r C a s e 1 = 9 C 1 N u m b e r O f S o l u t i o n s F o r C a s e 2 = 9 C 2 . . . N u m b e r O f S o l u t i o n s F o r C a s e 9 = 9 C 9 T o t a l N u m b e r O f S o l u t i o n s = 9 C 1 + 9 C 2 + . . . . . . . . . . . . . . . . . . . + 9 C 9 = 5 1 1
There is a way to get the value of 2 9 − 1 directly. Can you think of how to approach it this way?
Hint: The − 1 corresponds to the case where there is only 1 integer involved in the sum.
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( 1 9 ) + ( 2 9 ) . . . . . . ( 9 9 ) can be written as ( ( 0 9 ) + ( 1 9 ) + ( 2 9 ) + ( 3 9 ) . . . . . . . + ( 9 9 ) ) − ( 0 9 )
using binomial theorem
( ( 0 9 ) + ( 1 9 ) + ( 2 9 ) + ( 3 9 ) . . . . . . . + ( 9 9 ) ) = 2 9
and ( 0 9 ) = 1
so ,
( ( 0 9 ) + ( 1 9 ) + ( 2 9 ) + ( 3 9 ) . . . . . . . + ( 9 9 ) ) − ( 0 9 ) = 2 9 − 1
which gives
( 1 9 ) + ( 2 9 ) . . . . . . ( 9 9 ) = 2 9 − 1
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Not quite. You are explaining how to get from this sotluion to the value of 2 9 − 1 . See Happy Melodies' solution above.
Hi! Should we think about the faster method while solving the problem or should we think about it like how best to improve my solution after the problem is solved using the slower method ?
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When first solving a problem, any method works. Take the one that you're most comfortable with / the one that will give you the result. When writing a solution, it is useful to think about the best way to present it. The method here is fine, especially for those who do not have much experience with using different interpretations.
I was pointing out the possibility of another approach, because this was the only solution, at the point in time when I made my comment.
hey harsh i want to ask that the method i.e. ( n − 1 a − 1 ) is applicable to find all kind of integral solutions
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GJ The method is applicable only If x1,x2...xn are Natural Numbers.
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can u tell the number of integral solutions of following equation ( x 1 ) + ( x 2 ) + ( x 3 ) + ( 4 x 4 ) = 3 0 along with the solution
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@Rishabh Jain – Rishabh here we will have to take cases x 4 = 0 , x 4 = 1 , x 4 = 2 . . . . . . . . . . x 4 = 7 And then using the formula to find number of integral solutions for each case we add them
Reminds me of my first Torque Group note where I covered this idea only... click here
Motivations When we encounter such counting problems, we must first ask ourselves, is there a way to simplify the problem. And the first idea that should pop up in our mind is using balls and sticks.
Solution Notice the problem can be modelled as having 1 0 balls, then placing sticks in between any consecutive pairs of balls to indicate that there is a addition of the number of balls to the left of each stick and the remaining balls. For example, let 0 be balls and 1 be sticks. Then,
00010000000 represents 3 + 7 = 1 0 . And 000100100000 represents 3 + 2 + 5 = 1 0 .
Then, notice that there are 1 0 − 1 = 9 gaps between each consecutive pairs of balls, in which sticks can be placed. [Sticks cannot be placed on either end e.g. 10000000000, because the question states to use integers from 1 to 9 only, not 0 . ]
We have 2 choices for each of the 9 gaps, either to place a stick or not to place. Therefore, there are 2 9 ways to put sticks in the gaps. However, notice that we will have to subtract 1 case off, which is the case when no sticks are put at all [once again because 1 0 cannot be used, only integers 1 - 9 can be used.]
Hence, final answer is 2 9 − 1 = 5 1 1 .