A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions ?

Total number of questions = 10

Question in part A = 6

Question in part B = 7

Selecting to questions with at least 4 from each part A and part B, can from done in following way.

${}^6{C}_4{\times }^7{C}_6{\times }^6{C}_5{\times }^7{C}_5{\times }^6{C}_6{\times }^7{C}_4=(\frac{6!}{4!2!}\times \frac{7!}{6!1!})+(\frac{6!}{5!1!}\times \frac{7!}{5!2!})+\left(\frac{1\times 7!}{4!3!}\right)({\because }^n{C}_r=\frac{n!}{r!(n-r)!})=\left(\frac{6\times 5\times 7}{2}\right)+(\frac{6\times 7\times 6}{2}+(\frac{7\times 6\times 5}{3\times 2}=\left(105\right)+\left(126\right)+35$
266 ways