# choosing toys!

• September 27th 2012, 02:34 PM
jthomp18
choosing toys!
Sheila, a fifth grader, and Jimmy, a sixth grader, collect all their toys and decide to play together. Sheila has 18 different toys and Jimmy has 19 different toys. If Sheila and Jimmy decide to play with at least one toy from Sheila and one toy from Jimmy at a time, with how many different collections of toys (up to 37 toys at a time) can the both of them play?

1. 342

2. 116,964

3. More than 137 billion

4. 37C2
• September 27th 2012, 02:57 PM
MathoMan
Re: choosing toys!
Quote:

Originally Posted by jthomp18
Sheila, a fifth grader, and Jimmy, a sixth grader, collect all their toys and decide to play together. Sheila has 18 different toys and Jimmy has 19 different toys. If Sheila and Jimmy decide to play with at least one toy from Sheila and one toy from Jimmy at a time, with how many different collections of toys (up to 37 toys at a time) can the both of them play?

If I understand it correctly this means that for any number of toys (between 1 and 18) chosen among Sheila's toys any number of toys (between 1 and 19) can be chosen among Jimmy's toys, right!?
• September 27th 2012, 02:57 PM
MathoMan
Re: choosing toys!
$\left(\sum\limits_{k=1}^{18}{ 18 \choose k}\right) \cdot \left(\sum\limits_{j=1}^{19}{19 \choose j}\right)=137438167041$
• September 27th 2012, 03:29 PM
Plato
Re: choosing toys!
Quote:

Originally Posted by MathoMan
$\left(\sum\limits_{k=1}^{18}{ 18 \choose k}\right) \cdot \left(\sum\limits_{j=1}^{19}{19 \choose j}\right)=137438167041$

You can get the same answer from $\left( {2^{18} - 1} \right)\left( {2^{19} - 1} \right)$
• September 27th 2012, 03:31 PM
jthomp18
Re: choosing toys!
Thank you! That helps! The first answer you gave I had not seen before. The second one, I have. It makes sense now. thank you so much for all your help!!