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How many ways are there to distribute six indistinguishable balls into nine distinguishable bins?

The objective of this question is to find the number of ways the six indistinguishable balls can be distributed into nine distinguishable bins.

A mathematical method for determining the number of potential groupings in a set of objects in which the selection order becomes irrelevant is referred to as combination. The objects can be chosen in any order in combination. It is a set of $n$ items chosen $r$ at a time with no repetition. It is a type of permutation. As a result, the number of certain permutations is always greater than the number of combinations. This is the fundamental distinction between both.

Selections are another name for combinations being the classification of items from a certain set of items. The formula of combinations is utilized to quickly determine the number of distinct groups of $r$ items that can be constituted from the $n$ distinct objects present. To evaluate a combination, it is necessary to first understand how to compute a factorial. A factorial is referred to as the multiplication of all positive integers that are both less than and equal to the given number. The factorial of a number is denoted by an exclamation mark.

Expert Answer

The formula for the combination when the repetition is allowed is:

$C(n+r-1,r)=\dfrac{(n+r-1)!}{r!(n-1)!}$

Here $n=9$ and $r=6$, substituting the values in the above formla:

$C(9+6-1,6)=\dfrac{(9+6-1)!}{6!(9-1)!}$

$C(14,6)=\dfrac{(14)!}{6!(8)!}$

$=\dfrac{14\cdot 13\cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!}$

$C(14,6)=3003$

Example 1

Find the number of ways in which a team of $5$ players can be formed from a group of $7$ players.

Solution

Here, repetition of players is not allowed, therefore using the combination formula for no repetitions as:

${}^nC_r=\dfrac{n!}{r!(n-r)!}$

where,  $n=7$ and $r=5$ so that:

${}^7C_5=\dfrac{7!}{5!(7-5)!}$

${}^7C_5=\dfrac{7!}{5!2!}$

${}^7C_5=\dfrac{7\cdot 6 \cdot 5!}{2\cdot 5!}$

${}^7C_5=7\cdot 3$

${}^7C_5=21$

Example 2

$8$ points are chosen on a circle. Find the number of triangles having their edges at these points.

Solution

${}^nC_r=\dfrac{n!}{r!(n-r)!}$

where,  $n=8$ and $r=3$ so that:

${}^8C_3=\dfrac{8!}{3!(8-3)!}$

${}^8C_3=\dfrac{8!}{3!5!}$

${}^8C_3=\dfrac{8\cdot 7\cdot 6 \cdot 5!}{3\cdot 2\cdot 1\cdot 5!}$

${}^8C_3=8\cdot 7$

${}^8C_3=56$

Hence, there are $56$ triangles having their edges at $8$ points on a circle.

Example 3

Evaluate ${}^8C_3+{}^8C_2$.

Solution

Since ${}^nC_r \,+\, {}^nC_{r-1}={}^{n+1}C_{r}$.

$n=8$ and $r=3$, so the given question can be written as:

${}^8C_3\,+\,{}^8C_{3-1}={}^{8+1}C_{3}$

${}^8C_3\,+\,{}^8C_{3-1}={}^{9}C_{3}$

${}^{9}C_{3}=\dfrac{9!}{3!(9-3)!}$

${}^{9}C_{3}=\dfrac{9!}{3!6!}$

${}^{9}C_{3}=\dfrac{9\cdot 8\cdot 7\cdot 6!}{3\cdot 2\cdot 1\cdot 6!}$

${}^{9}C_{3}=84$

Or  ${}^8C_3\,+\,{}^8C_{3-1}=84$

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