#82
we divide a group of 6 kids in to 3 subgroups with 2 in each group. How many different ways?
first group has C(6,2) ways, second group has C(4,2) ways, third group has C(2,2) ways, there are 3! Ways to arrange the three groups, so total
C(6,2)*C(4,2)/3!= 15 ways
A1=45,A2=49,which number is in the sequence?
A3=A1+7=45+7=52
A4=A2+7=49+7=56..
It’s an arithmetic sequence with d=4
An=A1+(n-1)d=45+(n-1)*4=4n+41
633=4n+41, n is a positive integer, n=148th term in the sequence
Answer is 633