Question

A person has $12$ friends out of which $7$ are relatives. In how many ways can he invite $6$ friends if at least $4$ of them should be his relatives
A.$487$
B.$462$
C.$488$
D.$502$

Answer 1

At least four means the person has to invite a minimum $4$relatives it may be $5$or $6$. Make three cases according to the number of relatives and then use the permutation and combination to evaluate the total number of ways of inviting.

Complete step-by-step answer:
We are given that a person has $12$ friends out of which $7$ are relatives
We have to find out how many ways he can invite $6$ friends if at least $4$ of them should be his relatives.
First, we evaluate the number of friends.
That is, $12 - 7 = 5$
There will be three cases: first is when $4$ relatives are invited, second case is when $5$ relatives are invited and third case when $6$ relatives are invited.
Now we evaluate the number of ways in each case.
Case-$1$
When $4$relatives and $2$friends are invited
We can select $4$relatives out of $7$ relatives in ${}^7{C_4}$ways.
We can select $2$friends out of $5$ friends in ${}^5{C_2}$ways.
Therefore, total number of ways in this case is ${}^7{C_4} \times {}^5{C_2}$
Case-$2$
When $5$relatives and $1$ friend is invited
We can select $5$relatives out of $7$ relatives in ${}^7{C_5}$ways.
We can select $1$friend out of $5$ friends in ${}^5{C_1}$ways.
Therefore, total number of ways in this case is ${}^7{C_5} \times {}^5{C_1}$
Case-$3$
When $6$relatives and $0$ friend is invited
We can select $6$relatives out of $7$ relatives in ${}^7{C_6}$ways.
We can select $0$friends out of $5$ friends in ${}^5{C_0}$ways.
Therefore, total number of ways in this case is ${}^7{C_6} \times {}^5{C_0}$
Therefore, total number of ways will be the sum of all the tree cases
Write the total number of ways.
${}^7{C_4} \times {}^5{C_2} + {}^7{C_5} \times {}^5{C_1} + {}^7{C_6} \times {}^5{C_0}$
Use the formula ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$

$${}^7{C_4} \times {}^5{C_2} + {}^7{C_5} \times {}^5{C_1} + {}^7{C_6} \times {}^5{C_0} \\\ = \dfrac{{7!}}{{4!3!}} \times \dfrac{{5!}}{{2!3!}} + \dfrac{{7!}}{{5!2!}} \times \dfrac{{5!}}{{4!1!}} + \dfrac{{7!}}{{6!1!}} \times \dfrac{{5!}}{{0!5!}} \\\ = \dfrac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3 \times 2}} \times \dfrac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}} + \dfrac{{7 \times 6 \times 5!}}{{5! \times 2 \times 1}} \times \dfrac{{5 \times 4!}}{{4!}} + 7 \times 1 \\\ = 10(35) + 5(21) + 7 \\\ = 462 \\\$$

Therefore, total number of ways are $462$
Hence, option (B) is correct.

Note: In these types of questions analyse the number of cases properly according to the given condition in the question.
Don’t forget to add the number of ways of each case to get the total number of ways of selecting anything.