Question

A bag contains $23$ balls in which $7$ are identical. Then find the number of ways of selecting $12$ balls from bag.

• A. ${}^{18}{C}_7+{}^{18}{C}_8$
• B. ${}^{18}{C}_6+{}^{18}{C}_8$
• C. ${}^{18}{C}_{1}2+{}^{18}{C}_8$
• D. ${}^{18}{C}_{1}2+{}^{18}{C}_7$

Answer 1

The correct option is B ${}^{18}{C}_6+{}^{18}{C}_8$

Here, $n=23,p=7,r=12(r>p)$
$\therefore$ Required number of selections = $\sum _{r=5}^{12}{}^{16}{C}_r$
$={}^{16}{C}_5+{}^{16}{C}_6+{}^{16}{C}_7+{}^{16}{C}_8+{}^{16}{C}_9+{}^{16}{C}_{10}+{}^{16}{C}_{11}+{}^{16}{C}_{12}$
$=({}^{16}{C}_5+{}^{16}{C}_6)+({}^{16}{C}_7+{}^{16}{C}_8)+({}^{16}{C}_9+{}^{16}{C}_{10})+({}^{16}{C}_{11}+{}^{16}{C}_{12})$
$={}^{17}{C}_6+{}^{17}{C}_8+{}^{17}{C}_{10}+{}^{17}{C}_{12}(\because {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r)$
$={}^{17}{C}_{11}+{}^{17}{C}_9+{}^{17}{C}_{10}+{}^{17}{C}_{12}(\because {}^n{C}_r={}^n{C}_{n-r})$
$=({}^{17}{C}_9+{}^{17}{C}_{10})+({}^{17}{C}_{11}+{}^{17}{C}_{12})$
$={}^{18}{C}_{12}+{}^{18}{C}_{10}$
$={}^{18}{C}_6+{}^{18}{C}_8$