Question

There are total 6 questions in an MCQ examination.
First three questions have 4 choices.
Next three questions have 5 choices.
Find the total number of possible sequence of answers.

It's Permutations and Combinations question. Please answer.

Answer 1

$\large\underline{\sf{Solution-}}$

Since, it is given that first question have 4 choices. So, it means first question can be answer in 4 ways.

Again, it is given that second question have 4 choices. So, it means second question can be answer in 4 ways.

Also, it is given that third question have 4 choices. So, it means third question can be answer in 4 ways.

Now, further it is given that fourth question have 5 choices. So, it means fourth question can be answer in 5 ways.

Also, it is given that fifth question have 5 choices. So, it means fifth question can be answer in 5 ways.

Again, it is given that sixth question have 5 choices. So, it means sixth question can be answer in 5 ways.

So,

$\rm \: By\:fundamental\:principle\:of \: counting$

$\rm \: Number\:of\:ways \: to \: answer \: these \: questions \:$

$\rm \: = \: 4 \times 4 \times 4 \times 5 \times 5 \times 5$

$\rm \: = \: 20 \times 20 \times 20$

$\rm \: = \: 8000$

So, 8000 is the total number of possible sequence of answers.

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:^{n}P_{r} \: = \: \frac{n!}{(n - r)!} \: }}$

$\boxed{ \rm{ \:^{n}C_{r} \: = \: \frac{n!}{r! \: (n - r)!} \: }}$

$\boxed{ \rm{ \:^{n}P_{r} \: = \: r! \: \times \: ^{n}C_{r} \: }}$

$\boxed{ \rm{ \:^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r} \: }}$

$\boxed{ \rm{ \:^{n}C_{r} \: = \: ^{n}C_{n - r} \: }}$

$\boxed{ \rm{ \:^{n}C_{x} \: = \: ^{n}C_{y} \: \rm\implies \:x = y \: \: \: or \: \: \: n = x + y}}$