gmatdriller wrote:Hi Mitch,
How can we probably apply the separator method to the problem below?

Tanya prepared 4 different letters to be sent to 4 different addresses.
For each letter, she prepared an envelope with its correct address.
If the 4 letters are to be put into the 4 envelopes at random, what
is the probability that only one letter will be put into the envelope
with its correct address?
(A) 1/24 (B) 1/8 (C) 1/4 (D) 1/3 (E) 3/8

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040

GMATGuruNY wrote:

Chaitanya_1986 wrote: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

Use the separator method.

Let the 5-unit sum = DDDDD.

We can use these five D's to represent any integer in which the sum of the digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.

dhonu121 wrote:

GMATGuruNY wrote:

Chaitanya_1986 wrote: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

Use the separator method.

Let the 5-unit sum = DDDDD.

We can use these five D's to represent any integer in which the sum of the digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.

You did not count 221. and 23.
How did you account for these combinations ?

GMATGuruNY wrote:

Chaitanya_1986 wrote: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

Use the separator method.

Let the 5-unit sum = DDDDD.

We can use these five D's to represent any integer in which the sum of the digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.