Inspired by Brian Charlesworth
How many 3 digit numbers are there, such that the hundreds digit is smaller than or equal to the tens digit, which is smaller than or equal to the units digit?
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1 solution
In JEE there is a method called introducing dummy variables , These dummy variables bring equality among the digits,
let the three digit number be x y z provided x ≤ y ≤ z
Let the variables be a , b , where a brings equality between x and y and b brings equality between y and z
Now we have a set from which three objects have to be chosen. The set is { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , a , b } , no zero as already explained above by brian sir.
From this set we have to select 3 numbers so number of ways ( 1 1 3 )
For example if one selects 2 , a , 6 so there will be only one number following the given order , the number is 2 2 6 ,
another example if one selects 4 , a , b then the number is 4 4 4 .
its simply a method to solve such problems involving equalities and inequalities within seconds
⌣ ¨
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Perfect!
Does it work with 5 digit numbers?
How many 5 digit numbers are there, such that each successive digit exceeds or equal to its predecessor?
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Does your question mean that a b c d e f is a five digit number and given that a ≥ b ≥ c ≥ d ≥ e . Then the answer will be ( 5 1 4 ) − 1
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@Tanishq Varshney – Is it {1,2,3,4,5,6,7,8,9,a,b,c,d,e} and five objects to be chosen?
What does variables a,b,c,d,e implies to? Thanks.
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@Bk Lim – a b c d e f implies a five digit number , and the set is
{ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , w , x , y , z }
where w brings equality between a and b ; x brings equality between b and c , and so on for y and z .
suppose one selects 0 , y , 7 , 6 , z the number will be 7 6 0 0 0 ,
i edited my comment becoz if one takes 0 , x , y , z , w then a five digit number is not possible, we got to remove this case.
Hope it is understandable
There is a "one-line" solution to this problem.
Hint: Express 1 6 5 as a suitable ( r n ) .
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Oh, right. In general, we could look at this as a problem of placing n identical marbles into 9 distinguishable boxes, (representing the integers 1 through 9 . ) Then we have one-to-one correspondence between the desired n -digit numbers and non-negative integer solutions to the equation ∑ k = 1 9 a k = n , which is a stars and bars problem with solution
( n 9 + n − 1 ) = ( n n + 8 ) .
In this case n = 3 , so the solution is ( 3 1 1 ) = 1 6 5 .
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Yes, that's similar to what I was thinking.
Sir please be more explanatory in the line where you have used sigma ak = n
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@Space Sizzlers – Say we are looking for a 3 -digit number that meets the given conditions, and note that, given the conditions, none of the digits can be 0 .
Now let a i be the number of repetitions of the positive integer i , where 1 ≤ i ≤ 9 . Then any "satisfactory" 3 -digit number is represented by a non-negative integer solution to the equation
∑ i = 1 9 a i = a 1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 + a 9 = 3 .
This representation is unique since, for any 3 integers chosen, there is only one way to arrange them in ascending order. For example, if a 2 = 1 , a 5 = 2 , and all other a i 's are 0 then the unique, satisfactory 3 -digit number represented by this solution is 2 5 5 .
Since any solution to this equation corresponds to a unique satisfactory number, and any satisfactory number can be represented by a unique solution to the equation, we have established a one-to-one correspondence between these two sets, (i.e., the set of solutions to the equation and the set of all satisfactory numbers), and so the two sets have the same number of elements.
So finally, to answer this question, we just need to find the number of solutions to the equation we've established above. This is a stars and bars problem, which according to Theorem two in the link, with n = 3 , k = 9 , has solution
( 3 3 + 9 − 1 ) = ( 3 1 1 ) = 1 6 5 as found before.
This result can then be generalized to n -digit numbers.
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@Brian Charlesworth – Thank you sir for the kind reply to make a learner understand better (:
@Brian Charlesworth – I've spent some time on it and finally I understand a little bit.
It's like I place a line of numbers.
1 in front then follow by 11 seats, ready to fill in by 2,3,4,5,6,7,8,9, i.e. 11C3
Some of the results:
123x4567x89x represents 379
12xx345678x9 represents 228
123456xxx789 represents 666
Huh! Thank you so much!
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@Bk Lim – Yes, that's the idea. For more information, you can read up the wiki Stars and bars .
First note that since the hundreds digit cannot be 0 and must be less than or equal to the other digits, we know that the 3 -digit number cannot have any 0 's.
Now we can divide this into three cases:
(i) there are 9 numbers in which all three digits are the same value;
(ii) with three distinct digits chosen from the set A = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } there is only one way to arrange them in strictly ascending order. Such arrangements form a one-to-one correspondence with the 3 -digit numbers whose digits are distinct and in ascending order. This gives us ( 3 9 ) = 8 4 more numbers;
(iii) to count the specified 3 -digit numbers that are composed of only two distinct digits from A , we choose two digits in ( 2 9 ) = 3 6 ways and then choose one of these digits to duplicate. We can then again order the three digits in only one way, giving us 3 6 ∗ 2 = 7 2 more numbers.
The number of desired 3 -digit numbers is then 9 + 8 4 + 7 2 = 1 6 5 .