I came across a question that asked: Two punts can each hold 6 people. A party of 10 wishes to use these punts. In how many different ways can the party be divided? Assume that each member of the party is distinct.
Please could I check how to go about calculating the number of combinations?
Would I be correct if I say the answer is the following :
$^{10}C_6 + ^{10}C_5 + ^{10}C_4 = 672$