Maths doubts

classic Classic list List threaded Threaded
6 messages Options
Reply | Threaded
Open this post in threaded view
|

Maths doubts

Homer Simpson
This post was updated on .
Please explain how to do these. Thank you :)

1) At the end of the day, a bakery gives everything that is unsold to food banks for the needy. If it has 12 apple pies left at the end of a given day, in how many different ways can it distribute these pies among six food banks for the needy? Also, in how many different ways can the bakery distribute the 12 apple pies if each of the six food banks is to receive at least one pie?

2) A carton of 15 light bulbs contains one that is defective. In how many ways can an inspector choose 3 of the bulbs and get

a) one that is defective 91
b) non-defective ones 364
“Operator! Give me the number for 911!”
Reply | Threaded
Open this post in threaded view
|

Re: Maths doubts

Akshay Jain
1st apple pie can go to any1 of  the 6 food banks....similarly 2nd apple pie can go to any1 of 6 diff food banks....similarly 3rd and so on
so total ways =6*6*6*6...12 times=6^12
In the 2nd part...every bank must hav atleast 1 pie...so 1st of all giv 6 pies to 6 diff banks...since all apple pies are same this can b done only in 1 way
nkw distribute the remaining six in similar way discussed in 1st part of the ques
...6*6*...6 times=6^6 is the total no. Of ways
Akshay Jain
Masters in Economics
Delhi School of Economics
2013-15
Reply | Threaded
Open this post in threaded view
|

Re: Maths doubts

SINGHAM
1.If pies are indistinguishible, then
a) 17C5
b) If at least one should be given to each, 11C5


Reply | Threaded
Open this post in threaded view
|

Re: Maths doubts

SINGHAM
This post was updated on .
@Tuski, please let me know what do you mean by "not the ones that are not defective ". Thanks.

@Tuski, thanks for update.
Reply | Threaded
Open this post in threaded view
|

Re: Maths doubts

Ankur9
In reply to this post by Homer Simpson
Total bulbs = 15
Defective = 1
Non Defective = 14
a) Choose one defective bulb out of three

14C2 x 1C1 = 91

b) All are non defective

14C3 = 364
Reply | Threaded
Open this post in threaded view
|

Re: Maths doubts

Homer Simpson
In reply to this post by Homer Simpson
Thanks everyone. I got it now
“Operator! Give me the number for 911!”