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12 Feb 2021, 07:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (02:47) correct
45%
(02:47)
wrong
based on 121
sessions
History
Date
Time
Result
Not Attempted Yet
What is the number of permutation of the digits 1, 2, 3, 4 and 5 so that the sum of the digits at the first two places is smaller than the sum of the digit at the last two places ?
A. 46
B. 47
C. 48
D. 50
E. 60
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 46
B. 47
C. 48
D. 50
E. 60
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
11 Jun 2024, 05:26
Kudos
Bookmarks
The number of ways the sum of the first two digits exceeds that of the last two is the same as the number of ways the last two exceed the first two.
The total number of permutations is:
5! = 120
Now need to identify patterns where the sum of the first two EQUALS sum of the last.
15 42
23 41
43 52
The digits in each pair can be arranged 2 ways and the pairs also 2 ways for:
2*2*2 = 8 ways
Total ways where sums equal:
3*8 = 24
Subtract from 120 = 96
96 is the number of arrangements where the sums of the digits in first two/last two are unequal.
So 96/2 = 48 represents desired result
Posted from my mobile device
The total number of permutations is:
5! = 120
Now need to identify patterns where the sum of the first two EQUALS sum of the last.
15 42
23 41
43 52
The digits in each pair can be arranged 2 ways and the pairs also 2 ways for:
2*2*2 = 8 ways
Total ways where sums equal:
3*8 = 24
Subtract from 120 = 96
96 is the number of arrangements where the sums of the digits in first two/last two are unequal.
So 96/2 = 48 represents desired result
Posted from my mobile device
General Discussion
12 Feb 2021, 18:41
Kudos
Bookmarks
Bunuel
We can calculate each way and find the answer.
Let us check the last two digits pair wise
1) 4 and 5 : Sum =9
The first 3 digits can be anything => 3!*2=12
3! For first three digits and 2 for two ways in which 4 and 5 can be placed.
2) 3 and 5 : Sum =8
The first 3 digits can be anything => 3!*2=12
3! For first three digits and 2 for two ways in which 3 and 5 can be placed.
3) 2 and 5 : Sum =7
The first 2 digits can be 1 and 3 or 1 and 4=> 2(2!*2)=8
2! for ways within the first 2 digits and 2 for two ways in which 2 and 5 can be placed and for two sets of number 1&3 and 1&4
4) 1 and 5 : Sum =6
The first 2 digits can be 2 and 3 => (2!*2)=4
2! for ways within the first 2 digits and 2 for two ways in which 1 and 5 can be placed
5) 3 and 4 : Sum =7
The first 2 digits can be 1 and 2 or 1 and 5 => 2(2!*2)=8
2! for ways within the first 2 digits and 2 for two ways in which 3 and 4 can be placed and for two sets of number 1&2 and 1&5
6) 2 and 4 : Sum =6
The first 2 digits can be 1 and 3 => (2!*2)=4
2! for ways within the first 2 digits and 2 for two ways in which 2 and 4 can be placed
Total : 12+12+8+4+8+4=48
C
