Crazy is that !

Probability Level 4

If $x,y,z,a,$ and $b$ are non-negative integers, how many distinct 5-tuples $(x,y,z,a,b)$ that satisfy $x+y+z+a+b =15$ exist?

The answer is 3876.

2 solutions

Sujoy Roy
Nov 25, 2014

Total ways $=^{15+5-1}C_{5-1}=^{19}C_{4}=\boxed{3876}$

how and why? please explain

Mardokay Mosazghi - 6 years, 6 months ago

See Stars and bars .

Calvin Lin Staff - 6 years, 5 months ago

It is a formula. Given the equation:

$x_1 + x_2 + \ldots + x_n = r$

There are ${{n + r - 1} \choose {r}}$ different ways for the sum to be satisfied.

Sharky Kesa - 6 years, 6 months ago

Provided that ${x}_{1}, {x}_{2},...{x}_{n} \geq 0$ . And yes, nothing more special in it, it is just found by assuming that we have r balls which we have to 'distribute' in n boxes. And the no. of ways for the same is given by that formula which can be found easily.

Kartik Sharma - 6 years, 6 months ago

thanks for clarification.

Mardokay Mosazghi - 6 years, 6 months ago

Brock Brown
Dec 27, 2014

Solved with nested for loops:

count = 0
for a in xrange(16):
    for b in xrange(16):
        for c in xrange(16):
            for d in xrange(16):
                for e in xrange(16):
                    if a+b+c+d+e == 15:
                        count += 1
print count

Here's a slightly shorter solution:

from itertools import product
count = 0
for a,b,c,d,e in product(xrange(16), repeat = 5):
    if a+b+c+d+e == 15:
        count += 1
print count

Crazy is that !

The answer is 3876.

2 solutions

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