question_answer

We have 19 identical gems available with us which are needed to be distributed among A, B and C such that A always gets an even number of gems and a minimum of 2 gems. The number of ways this can be done is

A) \[120\]

B)        \[110\]

C) \[100\]

D)        \[90\]

Correct Answer: D

Solution :

[d] \[A+B+C=19\] Case I: If A = 2 then number of ways \[{{=}^{18}}{{C}_{1}}\] Case II: If \[A=4\]then number of ways \[{{=}^{16}}{{C}_{1}}\] Case IX: If \[A=18\]then number of ways \[{{=}^{2}}{{C}_{1}}\] \[\therefore \] Total number of ways \[{{=}^{18}}{{C}_{1}}{{+}^{16}}{{C}_{1}}+.....{{+}^{2}}{{C}_{1}}=90\]