Permutations And Combinations Question 150

Question: $ ^{14}C_4+\sum\limits _{j=1}^{4}{^{18-j}C_3} $ is equal to [EAMCET 1991]

Options:

A) $ ^{18}C_3 $

B) $ ^{18}C_4 $

C) $ ^{14}C_7 $

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ ^{14}C_4{{+}^{14}}C_3{{+}^{15}}C_3{{+}^{16}}C_3{{+}^{17}}C_3{{=}^{18}}C_4 $ .