Example:- Distribute 10 identical chocolate among 3 students such that each student gets none or more chocolate.
Solution:-
This type of problem we have studied earlier in word formation when we were to find total number of words from the letters of the word
MISSISSIPPI
n = 11 (total number of letters)
p = 4 (S)
q = 4 (I)
r = 2 (P)
∴ total number of words =
Here also the same condition, where
∴ total ways = Answer
This can be co-related with number of integral solution of equation.
In above question Let students be x, y & z
then
Here in place of ‘Lines(┃)’ we have used just ‘+’ symbol
To divide between 3 students we will be requiring 2 lines & in place of lines we have used ‘+’ sign.
∴ number of ways = number of solutions of above equation
= ¹²C₂
= Answer
(Q). In how many ways 5 identical chocolates can be distributed among 10 students if there is no restriction.
Solution:-
S₁ + S₂ + S₃ + S₄ + S₅ + S₆ + S₇ + S₈ + S9 + S₁₀ = 5
S₁, S₂, S₃, ……. S₁₀ ≥ 0
total ‘+’ sign = 9
Hence number of ways = ¹⁴C₉ = Answer
(Q). In how many ways 10 identical books can be distributed among 3 students if each student gets at least 1 book.
Solution:-
(Q). In how many ways 10 identical chocolates can be distributed among 4 children if a child has at most 1 chocolate.
Solution:-
0 Answer
(Q). In how many ways 4 identical chocolate can be distributed among 10 children such that a child gets at most 1 chocolate.
Solution:-
Simply we have to find number of ways of selecting 4 children out of 10 children.
= ¹⁰C₄ Answer
If chocolates are also 10 then number of ways = ¹⁰C₁₀ = 1 Answer
(Q). we have 30 identical pens. These are to be distributed among 4 students A, B, C, D such that student A gets at least 4 pens, student B gets none or more pens(Any nubmer of pen), student C gets at least 5 pens & student D gets at least 1 pen. Then in how many ways pen can be distributed.
Solution:-
now distribute 20 pens among A, B, C, D in anyway = ²³C₃ Answer
(Q). Find the number of terms in the expansion of (x + y + z)¹⁰⁰.
Solution:-
(x + y)² = x² + 2x¹y¹ + y²
= x²y⁰ + 2x¹y¹ +x⁰ y²
i.e. sum of powers of variable x, y in every term = 2
Here total number of terms = number of ways to distribute power 2 i.e. 2 identical objects between 2 person x, y in anyway.
⇒(x + y)³ = x³ +3x²y¹ +3x¹y² + y³
Sum of power of variable x, y in every term = 3
∴ Number of terms = number of ways to distribute power 3 i.e. 3 identical object between 2 person x, y in any way.
∴ (x + y + z)¹⁰⁰ ⇒ Total numbers of terms
x + y + z = 100
& condition is x≥0
y≥0
z≥0
= ¹⁰²C₂ Answer