300+ Math questions...Question #38

[CaptainOats]

hey guys,

I dont really get the answer to question no. 38 (especially the donut thing):


38. How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93



Solution:


Here, we have 4 digits (positive integer less than 10000 means that 9999 is the biggest and we can pretend that "5" is "0005") that must sum to 5.
Since we have 4 digits, we'll have 3 partitions. We're summing to 5, so we have 5 "donuts".
O O O O O

Since we can use 0, we can have multiple partitions in the same spot. For example, we could have:
|||OOOOO (which translates to 0005)
we could have:
||O|OOOO (which translates to 0014, or 14).

So, we view this as a permutation question: we have 8 total objects, 3 of which are identical to each other (the partitions) and 5 of which are identical to each other (the donuts). Using the permutation formula for which some objects are identical:
Total permutations = n!/r!s! = 8!/3!5! = 8*7*6/3*2*1 = 8*7 = 56... Choose (C).

[Bharat]

Answer: 56
Approach: 4 digit numbers formed using 0, 1, 2, 3, 4 & 5 such that total is of digits is exactly 5.
possible combinations leading to total of 5:
(5, 0, 0, 0) = 4C1 = 4 [combination for placing 5 anywhere in 4 places]
(4, 1, 0, 0) = 4C1 * 3C1 = 12 [combination for placing 4 & 1 anywhere in 4 places]
(3, 2, 0, 0) = 4C1 * 3C1 = 12 [combination for placing 3 & 2 anywhere in 4 places]
(3, 1, 1, 0) = 4C1 * 3C1 = 12 [combination for placing 3 & 0 anywhere in 4 places]
(2, 2, 1, 0) = 4C1 * 3C1 = 12 [combination for placing 1 & 0 anywhere in 4 places]
(2, 1, 1, 1) = 4C1 = 4 [combination for placing 2 anywhere in 4 places]

Total = 56

Let me know if you have questions. Also, I am interested in learning the donut method- appears to be a quick one. Thanks.