In this type of question wee will be required to find the value of Stirling Number of Second Kind.
Stirling Number of Second Kind:–
Formula for Stirling number of second kind is :-
= 63 + 3S(6, 2) + 9S(6, 3) + 4S(7, 3) + 16S(7, 4)
= 63 + 3(2⁶⁻¹ – 1) +9S(6, 3) + 4S(7, 3) + 16S(7, 4)
= 63 + 3×31 + 9[S(5, 2) + 3S(5, 3)] + 4[S(6, 2) + 3S(6, 3)] + 16[S(6, 3) + 4S(6, 4)]
= 156 + 9(2⁵⁻¹ – 1) +27S(5, 3) +4(2⁶⁻¹ – 1) + 12S(6, 3) + 16S(6, 3) + 64S(6, 4)
= 156 + 9×15 + 27S(5, 3) + 4×31 + 28S(6, 3) + 64S(6, 4)
= 415 + 27[S(4, 2) + 3S(4, 3)] + 28[S(5, 2) + 3S(5, 3)] + 64[S(5, 3) + 4S(5, 4)]
= 415 + 27[(2⁴⁻¹ – 1) + 3×⁴C₂] + 28[(2⁵⁻¹ – 1) + 3S(5, 3)] + 64[S(5, 3) + 4×⁵C₂]
= 415 + 27×7 + 81×6 + 28×15 + 84S(5, 3) + 64S(5, 3) + 64×4×10
= 4070 + 148S(5, 3)
= 4070 + 148[S(4, 2) + 3S(4, 3)]
= 4070 + 148[(2⁴⁻¹ – 1) + 3×⁴C₂]
4070 + 148[7 + 3×6]
= 4070 + 148×25
= 7770 Answer
Let us calculate the value of S(9, 4) using direct formula:-
Now come to question:-
(Q). In how many ways 7 people can be divided into 3 identical groups.
Solution:-
Since Groups are identical i.e. indistinguishable. So a particular object is in which group it does not matter. The only thing which matters is which objects are selected together.
∴ Total number of ways = 1 + 7 + 21 + 21 + 35 + 105 + 70 + 105
= 365 ways Answer
(Q). In the previous question if the condition is that no group should be empty i.e. all groups should have at least one object.
Solution:-
So we need to consider only 4ᵗʰ, 6ᵗʰ, 7ᵗʰ & 8ᵗʰ cases.
which gives → 21 + 105 + 70 + 105 = 301 ways Answer
⦿ These were the manual way to do these type of questions. Now we will use Stirling number to solve these type of problems:-
(Q). (previous question); In how many ways 7 people can be distributed into 3 identical groups anyway.
Solution:-
= S(7, 1) + S(7, 2) + S(7, 3)
↓
= 1 + (2⁷⁻¹ – 1) + S(7, 3)
= 64 + S(7, 3)
= 64 + [³C₀ 3⁷ – ³C₁ 2⁷ + ³C₂ 1⁷ ]
= 64 + [ 1×2187 – 3×128 + 3×1 ]
= 64 + ×1806
= 64 + 301
= 365 Answer
(Q). In how many ways 7 people can be distributed into 3 identical groups if each group gets at least 1 people(i.e. no group is empty).
Solution:-
S(7, 3) = 301 Answer
(Q). In how many ways 5 distinct balls can be distributed to 10 identical bags:-
(i) without any restriction
(ii) each bag gets atleast 1 ball
(iii) each bag gets at most 1 ball.
Solution:-
(ii) 0 Answer
(iii) 1 Answer