Test Index

NCERT Class XI Mathematics - Permutations and Combinations - Solutions

© examsnet.com
Question : 16 of 41
Marks: +1, -0
Find n if n1P3\,{}^{n-1}P_3 : nP4\,{}^{n}P_4 = 1 : 9
Solution:  
We have, n1P3\,{}^{n-1}P_3 : nP4\,{}^{n}P_4 = 1 : 9
(n1)!(n13)!\frac{(n-1)!}{(n-1-3)!} : n!(n4)!\frac{n!}{(n-4)!} = 1 : 9 ⇒ (n1)!(n4)!×(n4)!n!\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = 19\frac{1}{9}
(n1)!n(n1)!\frac{(n-1)!}{n(n-1)!} = 19\frac{1}{9}1n\frac{1}{n} = 19\frac{1}{9} ⇒ n = 9
© examsnet.com
Go to Question: