Question

in an examination , a candidate has to pass in each of the 5 subjects . in how many ways can he fail ?

Answer 1

answer :

the candidate can fail by falling in 1 or 2 or 3 or 4 or 5 subject out 5 in each case.

so ,

total number of ways in which he can fail.

5^ c 1 + 5^c 2 + 5^c3 + 5^c4 + 5^c5

5^c1 + 5^c2 + 5^c2 + 5^c1 + 5^c5 { n^C r = nCn - r}

= ( 5 + 10 + 10 + 5 +1)

= 31

be brainly

Answer 2

Answer:

The number of ways the candidate in 5 subjects is 31 ways.

Step-by-step explanation:

Given the question that the  candidate has to fail,

let the candidate fail in one subject then the possible ways is: $C_{1} ^{5}$

the candidate fails two subjects then the possible ways are: $C_{2} ^{5}$

let the candidate fail in three subjects then the possible ways are: $C_{3} ^{5}$

let the candidate fail in four subjects then the possible ways are: $C_{4} ^{5}$

let the candidate fail in five subjects then the possible ways are: $C_{5} ^{5}$

the total number of ways = $C_{1} ^{5} +C_{2} ^{5} +C_{3} ^{5} +C_{4} ^{5} +C_{5} ^{5}$

                                         = $\frac{5!}{(5-1)\!\times 1!} +\frac{5!}{(5-2)\!\times 2!}+ \frac{5!}{(5-3)\!\times 3!}+\frac{5!}{(5-4)\!\times 4!}+\frac{5!}{(5-5)!\times 5!}$

                                        = $\frac{5 \times 4!}{4!\times 1!} +\frac{5\times 4 \times3!}{3!\times 2!}+ \frac{5 \times 4 \times 3 \times 2!}{2!\times 3!}+\frac{5\times 4 !}{(5-4)\!\times 4!}+\frac{5!}{0!\times 5!}$

                                        = $\frac{5}{1!} +\frac{5\times 4 }{2!}+ \frac{5 \times 4 \times 3! }{2!\times 3 !}+\frac{5 }{1!}+\frac{5!}{ 5!}$

                                        = 5 + 10 + 10 + 5 + 1

                                        = 31

The number of ways the candidate in 5 subjects is 31 ways.