made easy test series

Question

What is the number of partition of X={a,b,c,d,e,f}.where 'a' and 'c' are always in same block?

Comments

i think u have missed somewhere

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yeah can you check??

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i think in the first solution u can check for your answer and match accordingly its the simplest way he has shown the solution.

Answer 1

Here, x={a,b,c,d,e,f} mean cardinality of set x= 6

when a and c occurs together means consider it as a single element.

so now the cardinality of the set becomes 5.

total number of partitions for a set having 5 elements = B₅(B= bell number)

and B₅=52.

https://en.wikipedia.org/wiki/Bell_number

 

Answer 2

In case you don't want to leave it as an application of some 'Bell Number', here's a more intuitive solution.

We can partition 5 symbols, ie {A, C}, B, D, E, F in 5 ways that is,

All of them in 1 partition

All of them divided in 2 partitions

Similarly 3, 4, 5 partitions

 

1 and 5 partitions:

 Either have all of them in one partition or each separately. Hence total = 2

 

2 partitions:

Can be divided as 2 elements in one partition and remaining 3 in the other OR 1 element in one partition and remaining 4 in the other. Ways of doings it = $\frac{5!}{3!2!} + \frac{5!}{1!4!} = 10+ 5= 15$

 

3 partitions:

Partitions can be formed in 2 ways. Either {1, 1, 3} or {1, 2, 2} . So ways of doing it = $\frac{5!}{3!2!} + \frac{5!}{2!2!2!} = 10+ 15= 25$

 

4 partitions:

Only one possible way to divide which is {1, 1, 1, 2}. Hence $\frac{5!}{1!1!1!2!3!} = 10$

 

Adding all of them =>

$2+15+25+10=52$

Comments

consider 2 partitions:

{a,c} {b,d,e,f}=1

{a,c,b}{d,e,f}     {a,c,d}{b,e,f}     {a,c,e}{b,d,f}     {a,c,f}{b,d,e}=4

{a,c,b,d}{e,f}   {a,c,b,e}{d,f}   {a,c,b,f}{d,e}   {a,c,d,e}{b,f}   {a,c,d,f}{b,e}   {a,c,e,f}{b,d}=6

{a,c,b,d,e}{f}   {a,c,b,d,f}{e}   {a,c,b,e,f}{d}   {a,c,d,e,f}{b}=4

Total=1+4+6+4=15!!! Can you explain how you got 10???

Answer 3

• Source : Wikipedia
• Answer is 52
•