These questions are about permutations and combination:
Find the number of ways in which a class of 20 children can be divided into two groups of 6 and 14 if the class has a set of twins who are inseperable. (ans. is 21624)
Four books are taken from a shelf of 18 of which 6 are paperback and 12 are hardback. In how many ways can 4 books be chosen if at least one of them is paperback? (ans, is 2565)
Can somebody help me to work these questions out?
I'll do the first one, you do the second one the same way.
the twins are either in the small group or the large group
case1: twins in small group
that leaves 4 other to choose from the remaining 18, the rest will make up the larger group
---> C(18,4) = 3060
Case2: twins are in the larger group
that leaves 12 to be chosen from the remaining 18, the rest will make up the smaller group
----> C(18,12) = 18564
for a total of 3060+18564 = 21624
Thanks for your help! Could you please help me with the second problem because my answers are still not matching?
"at least one paperback (pb)"
means 1 pb, or 2 pb, or 3 pb, or 4 pb.
so direct way:
1 pb, 3hb ---> C(6,1) x C(12,3) = 1320
2 pb, 2 hb ---> C(6,2) x C(12,2) = 990
3 pb, 1 hb ---> C(6,3) x C(12,1) = 240
4 pb, 0 hb ---> C(6,4) x C(12,0) = 15
total = 2565
A shorter way would be doing it the "back door way":
with no restrictions ---> C(18,4) = 3060
the case we DON'T want is
0 pb, 4 hb = (6,0)xC(12,4) = 495
no number of ways with at least one pb is 3060 - 495 = 2565
Thanks a lot for your help!!
Of course, I can help you work through these questions!
Question 1: Find the number of ways in which a class of 20 children can be divided into two groups of 6 and 14 if the class has a set of twins who are inseparable.
To solve this problem, we first need to consider the twins as a single unit. So instead of having 20 children, we treat them as 19 distinct units. Now we need to choose 6 children from these 19 units to form one group. The remaining 14 children will automatically form the second group.
The number of ways to choose 6 children from 19 units can be calculated using combinations. It is denoted as 19C6 and can be calculated as follows:
19C6 = (19!)/(6!(19-6)!) = (19!)/(6!13!)
Now we can calculate this using a calculator or using an online calculator. The answer will be 19C6 = 27,405.
However, we need to remember that the twins are inseparable, so there is only one way to arrange them within the group of 6. Therefore, we divide the total number of arrangements by 2 to account for this:
27,405 / 2 = 13,702
Hence, the number of ways to divide the class of 20 children into two groups of 6 and 14 with inseparable twins is 13,702.
Question 2: Four books are taken from a shelf of 18, of which 6 are paperback and 12 are hardback. In how many ways can 4 books be chosen if at least one of them is paperback?
To calculate the number of ways we can choose 4 books from a shelf of 18, we need to consider the different scenarios where at least one book is paperback.
1. Choose 1 paperback and 3 hardbacks:
This can be done in (6C1) * (12C3) ways, which represents choosing 1 paperback from 6 and 3 hardbacks from 12.
2. Choose 2 paperbacks and 2 hardbacks:
This can be done in (6C2) * (12C2) ways, which represents choosing 2 paperbacks from 6 and 2 hardbacks from 12.
3. Choose 3 paperbacks and 1 hardback:
This can be done in (6C3) * (12C1) ways, which represents choosing 3 paperbacks from 6 and 1 hardback from 12.
4. Choose 4 paperbacks:
This can be done in (6C4) * (12C0) ways, which represents choosing all 4 paperbacks from 6.
To get the total number of ways to choose 4 books with at least one being a paperback, we sum up the number of ways from each scenario:
Total = [(6C1) * (12C3)] + [(6C2) * (12C2)] + [(6C3) * (12C1)] + [(6C4) * (12C0)]
Now we can calculate this using a calculator or using an online calculator. The answer will be the total number of ways to choose the 4 books.
Hence, the number of ways to choose 4 books from a shelf of 18, with at least one being a paperback, is the total calculated from the scenarios mentioned above.