4 4 votes A child going on a trip is told that out of his 8 favorite toys, he can bring at most three toys. The number of ways he could select which toys he brings is ${ }_{8} P_{0}+{ }_{8} P_{1}+{ }_{8} P_{2}+{ }_{8} P_{3}$ ${ }_{8} C_{0}+{ }_{8} C_{1}+{ }_{8} C_{2}+{ }_{8} C_{3}$ ${ }_{8} C_{3}-\left({ }_{8} C_{0}+{ }_{8} C_{1}+{ }_{8} C_{2}\right)$ ${ }_{8} C_{0} \times{ }_{8} C_{1} \times{ }_{8} C_{2} \times{ }_{8} C_{3}$ Others goclasses2025-da-scholarship-test goclasses combinatory counting one-mark + – GO Classes 324 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote Very easy just select 0,1,2,3 given atmost 3..so let's stop at 3.. It is also mentioned just select so use combination.. 8c0 + 8c1 + 8c2 + 8c3 (B) is the answer. Luke Nebula answered Jul 28, 2024 Luke Nebula comment Share Follow 0 reply Please log in or register to add a comment.