Of $6$ applicants $3$ have backgrounds in business, $2$ have backgrounds in education and 1 has a background in recreation. Calculate the number of ways in which of the order of the $6$ interviews can be arranged when applicants having the same background are interviewed in succession?
What is wrong with my solution: $6P6- (3P3 \times 2P2 \times 1P1)$
On
First order the "backgrounds", then order the applicants "within each background".
This will give you $3!$ ways to order the backgrounds, for each of them $3!$ ways to order the applicants with background in business, $2!$ ways to order the applicants with background in education and $1!$ way of ordering the one applicant with background in recreation. The final result is (product rule): $3!\cdot 3!\cdot 2!\cdot 1!=72$.
Your result is incorrect because it just does not end up being equal to the right result ($72$), but as you have not stated how you got to your result, it is hard to tell where you've made a mistake.
What you did is subtract the number of ways of interviewing the three people who have backgrounds in business, the two people who have backgrounds in education, and the one person who has a background in recreation in that order from the total number of ways of interviewing six people. Therefore, you subtracted one of the admissible ways of interviewing the candidates from the number of ways the candidates could be interviewed without restriction.
What is to be found is the number of ways of arranging the six interviews when applicants having the same background are interviewed in succession. There are three backgrounds. The backgrounds can be arranged in $3!$ ways. The three people with business backgrounds can be interviewed in $3!$ orders, the two people with education backgrounds can be interviewed in $2!$ orders, and the remaining slot can be filled by the person with a background in recreation in one way. Therefore, the number of admissible arrangements of the interviews is $$3!3!2!1! = P(3, 3)P(3, 3)P(2, 2)P(1, 1)$$