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How many positive integers less than 10,000 are there in which the sum
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26 Sep 2022, 04:45
26 Sep 2022, 04:45
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Question Stats:
33% (03:26) correct
66% (02:54) wrong based on 15 sessions
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How many positive integers less than 10,000 are there in which the sum of the digits equals 5?
(A) 31
(B) 51
(C) 56
(D) 62
(E) 93
(A) 31
(B) 51
(C) 56
(D) 62
(E) 93
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GRE Math Essentials (2022)
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
Re: How many positive integers less than 10,000 are there in which the sum
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26 Sep 2022, 07:20
26 Sep 2022, 07:20
3
There could be six cases:
Case I:
The four digits consist of 1, 1, 1, 2.
Total numbers that can be formed using these four digits: 4. (2 can be in any of the four places)
Case 2:
1, 1, 3, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Case 3:
1, 2, 2, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Case 4:
1, ,4, 0, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Case 5:
0, 0, 0, 5
Total numbers that can be formed:4. (5 can be in any four places)
Case 6:
2, 3, 0, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Grand total of numbes: 4+12+12+12+4+12 = 56. Choice C.
Cheers.
Case I:
The four digits consist of 1, 1, 1, 2.
Total numbers that can be formed using these four digits: 4. (2 can be in any of the four places)
Case 2:
1, 1, 3, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Case 3:
1, 2, 2, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Case 4:
1, ,4, 0, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Case 5:
0, 0, 0, 5
Total numbers that can be formed:4. (5 can be in any four places)
Case 6:
2, 3, 0, 0
Total numbers that can be formed: 12. (4!/2, using Mississippi Rule of combination)
Grand total of numbes: 4+12+12+12+4+12 = 56. Choice C.
Cheers.
Re: How many positive integers less than 10,000 are there in which the sum
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22 Nov 2022, 23:16
22 Nov 2022, 23:16
2
1 digit:
5 so, 1 way
2 digits:
41-> 2 ways
32-> 2 ways
50-> 1 way
total 5 ways.
3 digits:
104-> 2*2*1 as (hundred's place 1,4 only allowed)(ten's place 2 possible)(one's place 1 possible)= 4 ways
113-> 3 ways as 3!/2! repeat
203-> 4 ways
221->3 ways
500-> 1 way
total: 15 ways
4 digits:
1004->6 ways as 2*3*2*1/2! (thousand's place 2 possible)(hundred's place 3 possible)(ten's 2) and 1 / repeat 2!
1013->9 ways
1022->9 ways
1112->4 ways
3200-> 6 ways
5000-> 1 way
total:35
So, for all digits 56 ways
5 so, 1 way
2 digits:
41-> 2 ways
32-> 2 ways
50-> 1 way
total 5 ways.
3 digits:
104-> 2*2*1 as (hundred's place 1,4 only allowed)(ten's place 2 possible)(one's place 1 possible)= 4 ways
113-> 3 ways as 3!/2! repeat
203-> 4 ways
221->3 ways
500-> 1 way
total: 15 ways
4 digits:
1004->6 ways as 2*3*2*1/2! (thousand's place 2 possible)(hundred's place 3 possible)(ten's 2) and 1 / repeat 2!
1013->9 ways
1022->9 ways
1112->4 ways
3200-> 6 ways
5000-> 1 way
total:35
So, for all digits 56 ways
