Coins

In a Bag contains 8 coins Gold, 7 Silver Coins, 4 Coins Copper and 3 Aluminum coins. How many different ways can remove 6 coins of this bag?


The answer is 70.

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3 solutions

Tejashwi Kumar
Apr 7, 2014

a + b + c + d = 6, d ≤ 3, c ≤ 4 C(9, 6) ways

When d > 3, C(5, 2) ways When c > 4, C(4, 1) ways

So, total = 84 - 14 = 70 ways

Can you explain the C(5,2) and C(4,1) part please?

Edwin Hughes - 6 years, 8 months ago

You want the coefficient of x 6 x^6

( n = 0 8 x n ) ( n = 0 7 x n ) ( n = 0 4 x n ) ( n = 0 3 x n ) \left(\sum _{n=0}^8 x^n\right) \left(\sum _{n=0}^7 x^n\right) \left(\sum _{n=0}^4 x^n\right) \left( \sum _{n=0}^3 x^n\right)

Aman Chandna
Feb 28, 2014

the following table gives you the answer let no. of aluminium coins drawn in a draw = A no. of copper coins drawn=c no. of silver coins=s no.of gold coins=g

we have a+C+s+g=6

a c s g 1 4 1 0\ 1 4 0 1\ 1 3 2 0\ 1 3 1 1\ 1 3 0 2\ 1 2 3 0\ 1 2 2 1\ 1 2 1 2\ 1 2 0 3\ 1 1 4 0\ 1 1 3 1\ 1 1 2 2\ 1 1 1 3\ 1 1 0 4\ 1 0 5 0\ 1 0 4 1\ 1 0 3 2\ 1 0 2 3\ 1 0 1 4\ 1 0 0 5\

0 4 2 0\ 0 4 1 1\ 0 4 0 2\ 0 3 3 0\ 0 3 2 1\ 0 3 1 2\ 0 3 0 3\ 0 2 4 0\ 0 2 3 1\ 0 2 2 2\ 0 2 1 3\ 0 2 0 4\ 0 1 5 0\ 0 1 4 1\ 0 1 3 2\ 0 1 2 3\ 0 1 1 4\ 0 1 0 5\ 0 0 6 0\ 0 0 5 1\ 0 0 4 2\ 0 0 3 3\ 0 0 2 4\ 0 0 1 5\ 0 0 0 6\ 2 4 0 0\ 2 3 1 0\ 2 3 0 1\ 2 2 2 0\ 2 2 1 0\ 2 2 0 2\ 2 1 3 0\ 2 1 2 1\ 2 1 1 2\ 2 1 0 3\ 2 0 4 0\ 2 0 3 1\ 2 0 2 2\ 2 0 1 3\ 2 0 0 4\ 3 3 0 0\ 3 2 1 0\ 3 2 0 1\ 3 1 2 0\ 3 1 1 1\ 3 1 0 2\ 3 0 3 0\ 3 0 2 1\ 3 0 1 2\ 3 0 0 3\

Omg! So complicated!

Pritipanna Ratha - 7 years, 3 months ago

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