How many positive integers less than 10,000 are there in

Pairs possible : 1,2,1,1 ; 1,3,1,0 ; 1,4,0,0 ; 0,1,2,2 ; 0,0,0,5 ; 0,0,2,3

which gives :
1,2,1,1 => 4!/3! = 4
0,0,0,5 =>=> 4!/3! = 4

1,3,1,0 => 4!/2! = 12
1,4,0,0 => 4!/2! = 12
0,1,2,2 => 4!/2! = 12
0,0,2,3 => 4!/2! = 12

Total = 4+4+ 12+12+12+12 = 56.

Bunnel I have solved it in this way but your methods seems to be quicker. But I couldn't understand.
Could you please explain in simple words.

This is correct. Also this is the best way to solve this question. +1. (Solved exactly the same way)

Answer: 56.

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)

What's the concept behind the separators?

This is what I found in the net about this question explaining the concept:

Hope it's clear.

this approach was more natural and my first idea to solve the promblem

Then I saw the Bunnel´s and AKProdigy87´s way to solve this problem.

The idea of Bunnel and AKProdigy87 method is that the sum of the digits must equal 5, and this five can be distributed among the 4 digits and numbers are made by "ones". Again, numbers are made by "ones"! please do not thing of numbers as 2, 3, 4 or 5 digit
for example:


|X|XX|XX = 0|1|2|2 = 0122
X|||XXXX = 1|0|0|4 = 1004

I hope it helps

thanks Bunuel
can u explain me this by using the formulae
How many positive integers less than 10,000 are there in which the sum of the digits equals 6?
thanks in advance

6 * (digits) and 3 ||| --> ******||| --> # of permutations of these symbols is \(\frac{9!}{6!3!}\).

Or: The total number of ways of dividing n identical items (6 *'s in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 6 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={6+4-1}_C_{4-1}={9}C3=\frac{9!}{6!3!}\).

Hope it's clear.

I don't like the sticks method. It is not intuitive at all. And I will no way even think of that in the exam. For me, the usual way is better.

Sum of digits --> 5 and Less than 10,000.

(0,0,0,5) - 4!/3! - 4
(0,0,1,4) - 4!/2! - 12
(0,0,2,3) - 4!/2! - 12
(0,1,1,3) - 4!/2! - 12
(0,1,2,2) - 4!/2! - 12
(1,1,1,2) - 4!/3! - 4

Add them all --> 56

Isnt this simple enough? And can be extrapolated easily to any question of this sort no?

If the sum of digits of a number is equal to 5, then each digit must be less than or equal to 5. Furthermore, we can consider each number as a “4-digit” number. For example, 5 can be considered as 0005 and 104 can be considered as 0104. We can classify the numbers into cases according to the largest digit of the number.

1) If the largest digit is 5, then the other three digits must be all 0. The number of ways to arrange one 5 and three 0’s is 4!/3! = 4.

2) If the largest digit is 4, then the other three digits must consist of one 1 and two 0’s. The number of ways to arrange one 4, one 1 and two 0’s is 4!/2! = 4 x 3 = 12.

3) If the largest digit is 3, then the other three digits must consist of one 2 and two 0’s OR two 1’s and one 0. The number of ways to arrange one 3, one 2 and two 0’s is 4!/2! = 12, and the number of ways to arrange one 3, two 1’s and one 0 is also 4!/2! = 12. So the total number of ways in this case is 12 + 12 = 24.

4) If the largest digit is 2, then the other three digits must consist of all 1 OR one 2, one 1 and one 0. The number of ways to arrange one 2 and three 1’s is 4!/3! = 4, and the number of ways to arrange two 2’s, one 1 and one 0 is also 4!/2! = 12. So the total number of ways in this case is 4 + 12 = 16.

Since the largest digit of the number can’t be 1 or 0, the total number of ways in which the sum of the digits is equal to 5 is 4 + 12 + 24 + 16 = 56.

Answer: C

CAN WE SOLVE THIS QUESTION WITH THE METHOD WE HAVE USED ABOVE?­A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

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d's we distribute in the original question are identical, whereas the individuals mentioned in the question you referenced are not. Hence, we must account for permutations of different individuals there.

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