Bangladesh Mathematical Olympiad 2023
Junior category
by Equation Tracker
Problem 4
Fuad has 6 pens. Among these, 3 pens are identical and other 3 pens are distinctly different.
How many ways Fuad can distribute these 6 pens among 3 people so that, everyone gets at
least one pen?
Solution. At first let’s distribute the pens among 3 people in such way, where some of them might
get 0 pen. To do
so, first, we distribute the 3 identical pens among them which can be done in total
5
of 3+3−1
=
= 10 ways (By Stars and Bars Theorem).
3−1
2
Then, we distribute the distinctlydifferent pens among 3 people in total of 33 = 27 ways.
So, total number of ways is 33 · 52 = 27 × 10 = 270.
Now, 270 is the number of ways to distribute pens where some might get no pen at all. So, we
want to count the number of ways where at least
one of the 3 people get 0 pen. First, choose any
3
1 people among the 3 which can be done in 1 ways. We
give him/her any pen. So, we can
won’t
3+2−1
3
distribute those 6 pens between other 2 people in 2−1 · 2 = 32 ways. So, The number of ways of
distribution when at least 1 gets 0 pen is 31 41 · 23 = 96 ways. But we have overcounted something.
Let the 3 people be M1 , M2 , M3 and the pens be A, A, A, B, C, D. Suppose we are not giving M3
any pens. Then we have considered the case,
(M1 , M2 ) −→ (0 pen, 6 pens) =⇒ (M1 , M2 , M3 ) −→ (0 pen, 6 pens, 0 pen)
Again, suppose we are not giving M1 any pens. Then we have considered the case,
(M3 , M2 ) −→ (0 pen, 6 pens) =⇒ (M1 , M2 , M3 ) −→ (0 pen, 6 pens, 0 pen)
This is clearly an overcounting. So, considering all cases, we get total 3 overcounting.
5
3 4
3
Therefore, final answer is
·3 −
· 23 + 3 = 177 .
3
1 1
Lemma: Distribution of Objects
We can distribute m identical and n distinct object in k ≥ 3 distinct boxes (such that no boxes
are empty) in total
k−2 X
m+k−1
k
m+k−2
k
n
n
·k −
· (k − 1) +
(k − 1 − i)
k−1
1
k−2
i
i=1
ways. If k = 2,
m+k−1
k
m+k−2
n
·k −
· (k − 1)n
k−1
1
k−2
1