Bird Problem

A: Report includes 2 birds

Choose 2 out of the 6. 6C2 = 6!/4!2! = 15


B: Report includes 3 birds

Choose 3 out of 6. 6C3 = 6!/3!3! = 20


Probability of A or B = A + B = 15 + 20 = 35.

itsratul wrote:A second grade class is writing reports on birds. The students’ teacher has given them a list of 6 birds they can choose to write about. If Lizzie wants to write a report that includes two or three of the birds, how many different reports can she write?

Ans: 35

if she wants to write report about 2 bird she can choose this two bird
first she has 6 bird to choose from in second time she has five bird to choose 6*5=30
if she wants to write report about 3 bird she can choose this 3 bird
first she has 6 bird to choose from in second time she has five bird to choose third time she has 4 bird to choose from 6*5*4=120
so total 30+120=150

total ways to write report choosing 2 birds

6C2

total ways to write report choosing 3 birds

6C3

total ways = 6C2+ 6C3 = 15 + 20 = 35 (answer)

amising6 wrote:

itsratul wrote:A second grade class is writing reports on birds. The students’ teacher has given them a list of 6 birds they can choose to write about. If Lizzie wants to write a report that includes two or three of the birds, how many different reports can she write?

Ans: 35

if she wants to write report about 2 bird she can choose this two bird
first she has 6 bird to choose from in second time she has five bird to choose 6*5=30
if she wants to write report about 3 bird she can choose this 3 bird
first she has 6 bird to choose from in second time she has five bird to choose third time she has 4 bird to choose from 6*5*4=120
so total 30+120=150

Hi,

order doesn't matter here. A report on birds AB is not different from a report on birds BA.

For two birds, 6*5, because there are 2! different ways of ordering and because we want to remove order, we would divide 6*5 by 2!, giving us 15.

Similarly, for three birds, 6*5*4, because there are 3! ways of ordering the birds and because order doesn't matter, we would divide 6*5*4 by 3!, giving us 20.

Of course, we can also solve by explicitly using the combinations formula as albatross and indiantiger did.

albatross86 wrote:A: Report includes 2 birds

Choose 2 out of the 6. 6C2 = 6!/4!2! = 15


B: Report includes 3 birds

Choose 3 out of 6. 6C3 = 6!/3!3! = 20


Probability of A or B = A + B = 15 + 20 = 35.

Just a quick question here.

If Lizzie chooses to write about 2 birds first (6 C 2), the remaining no. of birds is 4. She can choose 3 birds out of these 4 in 4 C 3 ways? So shouldn't the answer be 6 C 2 + 4 C 3 = 12 + 4 = 16???

r2kins wrote:

albatross86 wrote:A: Report includes 2 birds

Choose 2 out of the 6. 6C2 = 6!/4!2! = 15


B: Report includes 3 birds

Choose 3 out of 6. 6C3 = 6!/3!3! = 20


Probability of A or B = A + B = 15 + 20 = 35.

Just a quick question here.

If Lizzie chooses to write about 2 birds first (6 C 2), the remaining no. of birds is 4. She can choose 3 birds out of these 4 in 4 C 3 ways? So shouldn't the answer be 6 C 2 + 4 C 3 = 12 + 4 = 16???

Hi,

The wording of the question is important. Because the question says 2 reports OR 3 reports, Lizzie writing 2 or 3 reports are INDEPENDENT scenarios, and order is not a concern. She can write 10 different reports containing 2 birds, and 15 different reports containing 3 birds. Thus, the number of ways she can "write a report that includes two or three of the birds" is 35.

chossing two birds out of 6=6C2=15
chossing 3 birds out of 6=6C3=20
total=20+15=35