Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.
See classesLearn live with an expert or move at your own pace. Every option includes the complete TTP study system.
Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.
with Josh Braslow, 20+ years in test prep
Self-paced EA prep. Study on your schedule.
with
Complete access from day one. Study on your schedule.
Although your solution is correct.. the example that you gave is I think a typing mistake..sureshbala wrote:Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.
For example, if we fix the digits 1,2 and 3 the arrangements possible are
1,2,3, 6.4,5
1,2,3, 6,5,4
i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.
Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
Yup...that's a typo...thanks for the correctiondumb.doofus wrote:Although your solution is correct.. the example that you gave is I think a typing mistake..sureshbala wrote:Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.
For example, if we fix the digits 1,2 and 3 the arrangements possible are
1,2,3, 6.4,5
1,2,3, 6,5,4
i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.
Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
1) 1,2,3,6,4,5
2) 1,2,3,5,6,4 (This one your typed incorrectly)
