There is a young boy's birthday party which 3 friends have attended. The mother has arranged 10 games where a prize is awarded for winning a game. The prizes are identical. If each of the 4 children receives at least one prize, then how many distributions of prizes are possible?
84
Understanding the Prize Distribution Problem
This problem involves finding the number of ways to distribute identical prizes to distinct children, with the condition that every child receives at least one prize. This is a common type of combinatorics problem.
Problem Setup
- Number of identical prizes (games): $n=10$
- Number of distinct children (boy + friends): $k=4$
- Condition: Each child must receive at least one prize.
Applying the Constraint
Since each of the 4 children must receive at least one prize, we can start by giving one prize to each child. This uses up 4 prizes.
Remaining prizes = Total prizes - Prizes given initially
Remaining prizes = $10 - 4 = 6$.
Now, we need to distribute these remaining 6 identical prizes among the 4 distinct children. There are no further restrictions, meaning a child can receive zero or more additional prizes.
Using the Stars and Bars Method
This scenario is equivalent to finding the number of non-negative integer solutions to the equation:
$x_1 + x_2 + x_3 + x_4 = 6$
where $x_i$ represents the number of *additional* prizes received by the $i$-th child.
The "stars and bars" formula for distributing $n'$ identical items into $k$ distinct bins is:
$$ \binom{n' + k - 1}{k - 1} $$In our case:
- $n' = 6$ (the remaining identical prizes, or "stars")
- $k = 4$ (the distinct children, needing $k-1$ "bars" to divide them)
Calculation
Plugging the values into the formula:
Number of distributions = $ \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} $
Now, we calculate the binomial coefficient:
$$ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7 \times 6!}{ (3 \times 2 \times 1) \times 6!} $$Cancel out $6!$:
$$ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} $$Simplify the calculation:
$$ \binom{9}{3} = \frac{504}{6} = 84 $$Conclusion
Therefore, there are 84 possible ways to distribute the 10 identical prizes among the 4 children such that each child receives at least one prize.
Important Questions from Permutations and Combinations
Three boxes are coloured red, blue and green and so are three balls. In how many ways can one put the balls one in each box such that no ball goes into the box of its own colour?
Six indistinguishable balls are to be distributed amongst A, B and C, such that each gets at least one. Then the number of ways to make this distribution is
A palindrome is a word that reads the same backwards and forwards. For example, the word LEVEL is a palindrome. If we are allowed to construct words that need not have a meaning, then in how many different ways can we construct a five-letter palindrome using English alphabets?
If nPr = 720 and nCr = 120, then the value of r is:
What will be the number of permutations of n different things, taken r at a time, where repeatition is allowed?
