Counting/ Combinatorics Problem

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harrybm: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Tue Jan 03, 2012 7:29 am

by harrybm » Fri Jan 13, 2012 12:35 am

Can Anyone help me with this Problem:

How many Positive integers less than 10000 are there, in which the sum of the digit equal to 5?
a) 31
b) 51
c) 56
d) 62
e) 93

Can anyone help pleasee??

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Source: Beat The GMAT — Problem Solving | Fri Jan 13, 2012 12:35 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Fri Jan 13, 2012 12:41 am

harrybm wrote:Can Anyone help me with this Problem:

How many Positive integers less than 10000 are there, in which the sum of the digit equal to 5?
a) 31
b) 51
c) 56
d) 62
e) 93

Can anyone help pleasee??

We need to find integers between 0 to 9999, in which the sum of digits adds up to 5.
(1) One digit is 5 and all other are 0: 0005, 0050, 0500, 5000 or we can say that no. of ways we can arrange the digits = 4!/3! = 4 ways
(2)Three 1's and one 2: 1112, this can be done in 4!/3! = 4 ways
(3) One 4 and one 1: 4100, this can be done in 4!/2! = 12 ways
(4) One 3 and one 2: 3200, this can be done in 4!/2! = 12 ways
(5) One 3 and two 1's: 3110, this can be done in 4!/2! = 12 ways
(6) Two 2's and One 1: 2210, this can be done in 4!/2! = 12 ways

Therefore, required number of positive integers = (12 * 4) + (4 * 2) = 48 + 8 = 56

The correct answer is C.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jan 13, 2012 5:22 am

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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Fri Jan 13, 2012 12:39 pm

Groups:
(0,0,1,4)
(0,0,2,3)
(0,1,1,3)
(0,1,2,2)
for the above 4 groups = 4* 4!/2!=48
======
(0,0,0,5)
(1,1,1,2)
for the above 2 groups = 2* 4!/3!=8

Total=48+8+56

(C) is the answer

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harrybm: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Tue Jan 03, 2012 7:29 am

by harrybm » Fri Jan 13, 2012 6:44 pm

Anurag@Gurome wrote:
harrybm wrote:Can Anyone help me with this Problem:

How many Positive integers less than 10000 are there, in which the sum of the digit equal to 5?
a) 31
b) 51
c) 56
d) 62
e) 93

Can anyone help pleasee??
We need to find integers between 0 to 9999, in which the sum of digits adds up to 5.
(1) One digit is 5 and all other are 0: 0005, 0050, 0500, 5000 or we can say that no. of ways we can arrange the digits = 4!/3! = 4 ways
(2)Three 1's and one 2: 1112, this can be done in 4!/3! = 4 ways
(3) One 4 and one 1: 4100, this can be done in 4!/2! = 12 ways
(4) One 3 and one 2: 3200, this can be done in 4!/2! = 12 ways
(5) One 3 and two 1's: 3110, this can be done in 4!/2! = 12 ways
(6) Two 2's and One 1: 2210, this can be done in 4!/2! = 12 ways

Therefore, required number of positive integers = (12 * 4) + (4 * 2) = 48 + 8 = 56

The correct answer is C.

Thank you Anurag, GMATGuruNY and AnurangSahu for the comprehensive explanation. However I'm always confused every time I see tough counting problems, especially to figuring out the approach in the beginning? Do you have any tips on this? (maybe any method to cluster problems on specific type, etc?)

Moreover I have further question in related topic:

There are 6 people sitting together, A, B, C, D, E, F. and A doesn't want to sit with B. How many arrangement can be made?

1) I understand I can solve like this:
Total Outcome - total outcome of 5 arrangement x 2 (since A & B interchangeable)
6! - (5! x 2) = 480 arrangement

But why cant I solve it like this:

6! - 6!/2! ?
6! is the total outcome
6!/2! is the total outcome considering 2 of them are interchangeable like word: PIZZA, since Z is interchangeable there are 5!/2! = 60 outcomes.

Please advice where do I go wrong here? and is there any alternative approach beside the first method? Thank you before

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ame: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat Jan 21, 2012 2:41 am

by ame » Sat Jan 21, 2012 2:48 am

here , the problem is same as solving the equation x1+x2+x3+x4=5.
where x1,x2,x3,x4 can contain intergers (0,1,...,9)

as we have to find the number of integers less than 10000, whose digit's sum adds upto 5. So we can compare 4 distinguisable boxes where each box can have any integer between 0to9.
finding the coeffecient of x^5in the equation (1+X+X^2+X^3+...+X^9)^5
i.e. 56.

Ans.56[/img]

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