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A pencil case can hold seven pens and eight pencils. Meg chooses either a pen or pencil. Then John is to choose a pen and a pencil. In which case (based on Megs initial choice) would John have to have the greater number of choices.
Does this look right?
If Meg chooses pen:
7C1x6C1x8C1
=336
If Meg chooses pencil:
8C1x7C1x7C1
=392
Therefore, John will have more choices if Meg chooses a pencil.
Last edited by CroatBoy (2009-10-05 12:30:46)
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Hello, CroatBoy!
A pencil case can hold seven pens and eight pencils.
Meg chooses either a pen or a pencil.
Then John is to choose a pen and a pencil.
In which case (based on Megs initial choice) would John have to have the greater number of choices?Does this look right?
If Meg chooses pen:
7C1 x 6C1 x 8C1 ?
=336If Meg chooses pencil:
8C1x7C1x7C1 ?
=392Therefore, John will have more choices if Meg chooses a pencil.
Right answer . . . for the wrong reasons
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