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Carol is buying birthday gifts for her niece. She has a list of 10 potential gifts from which to choose, but she can

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Carol is buying birthday gifts for her niece. She has a list of 10 potential gifts from which to choose, but she can

by BTGmoderatorDC » Mon Dec 28, 2020 5:30 pm

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Carol is buying birthday gifts for her niece. She has a list of 10 potential gifts from which to choose, but she can only afford to buy 2 gifts. How many different pairs of gifts can Carol buy?

A. 10
B. 20
C. 45
D. 90
E. 200


OA C

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Re: Carol is buying birthday gifts for her niece. She has a list of 10 potential gifts from which to choose, but she can

by Scott@TargetTestPrep » Fri Jan 01, 2021 5:10 am
BTGmoderatorDC wrote: ↑
Mon Dec 28, 2020 5:30 pm
 Carol is buying birthday gifts for her niece. She has a list of 10 potential gifts from which to choose, but she can only afford to buy 2 gifts. How many different pairs of gifts can Carol buy?

A. 10
B. 20
C. 45
D. 90
E. 200


OA C

Solution:

The number of ways Carol can choose to buy 2 gifts from a list of 10 (where order doesn’t matter) is 10C2 = (10 x 9)/2 = 45.

Answer: C

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Re: Carol is buying birthday gifts for her niece. She has a list of 10 potential gifts from which to choose, but she can

by deloitte247 » Sat Jan 02, 2021 12:26 pm
$$Total\ potential\ gifts\ =10\ gifts$$
$$Carol\ cano\ nly\ afford\ to\ buy=2\ gifts$$
$$So,\ \ Carol\ will\ be\ selecting\ 2\ out\ of\ 10\ gifts=>$$
$$10C2\ which\ is\ in\ forn\ nCr$$
$$=>\ \frac{n!}{r!\cdot\left(n-r\right)!}$$
$$=>\frac{10!}{2!\cdot\left(10-2\right)!}$$
$$=>\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$=>\frac{10\cdot9}{2\cdot1}$$
$$=>\frac{90}{2}$$
$$=>45$$
$$Answer\ =\ C$$
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