# Partitioning – some examples

## 14 thoughts on “Partitioning – some examples”

1. Vaibhav Jain |

find the number of integral solution |x|+|y|+|z|=15

a.900 b.901 c.902 d.903

How would be approach this Sir ?

2. Sir can you please share the approach for the above problem?

• There is a standard formula (can’t remember it offhand), else it is a long and painful question unless you can visualise it as a 3-D Coordinate Geometry question. Certainly not of CAT scope…I would leave it if faced with it in an exam. Not worth the trouble.

regards
J

3. Gaurav Raj |

Sir, for the 1st question when it is said that he bought 3 types of fruits comprising of apple banana and chickoo,doesn’t this mean that he would have bought at least 1 of each of these

4. Gaurav, not necessarily (though I agree it could be interpreted in that way). For example, if we said “in your second year of MBA, you will be required to take 15 electives, across marketing, finance, HR, strategy and operations”, it doesn’t mean that you necessarily need all these fields to be represented in your selection. This was the way in which I intended it to be interpreted. Anyway, apologies for the ambiguity ðŸ™‚ I hope the concept is clear nonetheless?

regards
J

5. Aashika Aggarwal |

sir in the above question kitkat wala.. agar wahan 40 ki jagah upto 40 kitkats hota to hum dummy person include krte na? nd tab answer wud hav been 24C8 ways right???

6. debabrata |

sir, for the above problem the ans is coming 902, can u pls tell whether it is right or wrong?
|x|+|y|+|z|=15, now
case 1:none of them can be zero
no of ways = 14c2 *(2^3)=728, 2^3 because each can be +ve or -ve

case 2: two zero out of three
no of ways =3*2=6

case 3: one zero out of three
no of ways=14c1*(2^2)*3=168

thus ans =728+168+6=902

7. seema |

sir, In positive solutions for x+y+z=20 , how did you calculated it to be 19C2 , plz explain (where this 19 came from)?

• From the previous posts ðŸ˜› Partitioning with no group empty => (n-1)C(r-1). Since we want positive solutions, 0 nahin ho sakta and so group cannot be empty.

regards
J

8. Lokesh |

Sir, I’ve been following the way you emphasize logic needs to be hold up over formulas, but can’t seem to understand why it works that way for n-r+1Cr-1. For the left over objects, how did we arrive at this way to distribute them amongst the group?

9. Rohit Gupta |

Sir, got everything clear and I was solving side by side as well, tho got one wrong. In that jacob had 3, I did 3 to jacob and rest 5 should have non empty solutions so 8c4. Where am I going wrong, Sir?

• Jacob has “at least” 3. You have given him “exactly 3” . He could get more as well…

regards
J