Home > Practice > Arithmetic Aptitude > Permutations and Combinations > Miscellaneous
1. If find n?
(n+1)P3 = (n+1) × n × (n–1)
Now,
5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1)
Or, 5(n−2) = 4(n+1)
Or, 5n − 10 = 4n + 4
Or, 5n − 4n = 4 + 10
Hence, n = 14
Note:nPr =
Solution:
nP3 = n × (n–1) × (n–2)(n+1)P3 = (n+1) × n × (n–1)
Now,
5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1)
Or, 5(n−2) = 4(n+1)
Or, 5n − 10 = 4n + 4
Or, 5n − 4n = 4 + 10
Hence, n = 14
Note:nPr =