Question

The number of selecting at least 4 candidates from 8 candidates is
A. 270
B. 70
C. 163
D. None of these

Answer 1

In the above question they have written that total 8 candidates are there in that number of selecting the candidate at least that is$\,r\ge 4$. Formula which is used for this problem is $^{n}{{c}_{r}}=\dfrac{n!}{r!(n-r)!}$ where n is the total number of candidates and r is the selecting the candidate from total number of candidates.

Complete step by step answer:
According to question there are 8 candidates out of which 4 are selected that is n =total number of candidates and r is selecting the candidates hence total number of ways is $^{n}{{c}_{r}}$
After substituting the value of $\,n=8$and $\,r\ge 4$(at least 4 candidates are selected).
Formula which is used here is ${{\,}^{n}}{{c}_{r}}=\dfrac{n!}{r!(n-r)!}$
$\therefore$Total number of ways ${{=}^{8}}{{c}_{4}}{{+}^{8}}{{c}_{5}}{{+}^{8}}{{c}_{6}}{{+}^{8}}{{c}_{7}}{{+}^{8}}{{c}_{8}}---(1)$
$^{8}{{c}_{4}}=\dfrac{8!}{4!(8-4)!}$
After simplifying further,
$^{8}{{c}_{4}}=\dfrac{8!}{4!(8-4)!}=70---(2)$
Similarly we can find,
$^{8}{{c}_{5}}=\dfrac{8!}{5!(8-5)!}=56---(3)$
$^{8}{{c}_{6}}=\dfrac{8!}{6!(8-6)!}=28---(4)$
$^{8}{{c}_{7}}=\dfrac{8!}{7!(8-7)!}=8---(5)$ $^{8}{{c}_{8}}=\dfrac{8!}{8!(8-8)!}=1---(6)$
After substituting the value of equation $(2),(3),(4),(5),(6)$ in equation$(1)$.
Total number of ways $=\dfrac{8!}{4!(8-4)!}+\dfrac{8!}{5!(8-5)!}+\dfrac{8!}{6!(8-6)!}+\dfrac{8!}{7!(8-7)!}+\dfrac{8!}{8!(8-8)!}$
Total number of ways $=70+56+28+8+1$
Total number of ways $=163$

So, the correct answer is “Option C”.

Note: According to given question here there are selecting at least 4 candidates which is $\,r\ge 4$
If the question is asked there are selecting 4 candidates then that means r is exactly 4 that is $r=4$
So, in this case we have to find only ${{\,}^{n}}{{c}_{r}}$ by using the formula and substitute the value$\,\,\,r=4$. Total number of ways will be$70$. So in this way problems can be solved in a similar manner.