# PnC Examples 4

## 21 thoughts on “PnC Examples 4”

In the first question , why we are not taking repeated letters in arrangements , as in question its stated that all possible arrangements ….

2. If I say all possible arrangements of 5 people in 5 seats, will you put the same person in all 5 seats? No, right? All possible arrangements, by default, means no repetition. (In fact as I mentioned in an earlier post on PnC, both the nPr and nCr formulae are valid only when there is a non repetition (one to one) scenario.

Thus for example the number of ways to arrange the word CAT would be 6. ACT, ATC, CAT, CTA, TAC, TCA. One could not take AAC or TTT or any such as an arrangement of the letters of CAT.

regards
J

Thanks …i understood now ..

4. sharmi |

hi, in case of NONAGON i am getting 206.Starting with A=6!/(2!3!).Starting with G=6!/(2!3!) .starting with NA=5!/(2!2!).with NG=5!/(2!2!).with NOA=4!/2!.with NOG=4!/2!.then NONAGNO.then NONAGON.hence 206.where am i going wrong??

5. Sharmi, have you considered cases starting with NN? There will be 5!/2! = 60 cases there…

6. pragdeesh |

I can’t even understand the first question if you don’t mind can you please explain me that “triangle” question?

• No, if I explain in any more detail the point is lost. You have to think, break your head on it, else you will only know a formula but never understand how to apply it. No value.

regards
J

7. pragdeesh |

dude i got triangle –check my understanding is weather right.
to get T as is we have 2 complete a,e,g,i,l,n, and r–7 which has 2 be arranged in 7! ways-7*7!
then 2nd letter R to have R in second place we have complete a,e,g,i,l & n-6 in 6! ways-6*6!
3rd letter i for this we have a,e,g in front of Iand 5 other word so 3 *5! ways
4th …A since a is first alphabet we don’t have any other letter in front of A so 0 and we have 4 other letters -0*4! ways
5th ,it is N we have e,g,l in front of N and we have 3 other words to be arrange to complete the process.-3*3!
then for 6th letter G we have e in front of g so 1*2!
then last 1*1!,,,,,(7*7!+6*6!+3*5!+0*4!+3*3!+1*2!+1*1!)
FOR nonagon
AGNNNOO
1st–(n)- 2* (6!/3!*2!)—120
2nd-(o)-3*(5!/3!*2!)—-30
3rd-(n) -2*(4!/2!*2!)–12
4th-(a)-0*(3!/2!.1!)
5th-(G)-0*(2!/2!)
6th-(o)-1*(1!)

i am making mistake can you correct me ?

8. pragdeesh |

NONAGON
agnnnoo-2*(6!/3!*2!)-120 ways
nnnagoo-2*(5!/2!)—120 ways
nooagnn-2*(4!/2!)—24 ways
nonnago-0*(3!)–0 ways
nonagno-0*(2!)-0 ways
nonagon-1*1!- 1 ways
total——265 ways
it will occur in 266 is my procedure is right?

• If it weren’t, there would be no chance of getting the correct answer in something like this ðŸ˜›

Also, try and struggle for an answer rather than asking for help. Else there is no way you will improve. Whatever ability and agility I have in maths is largely due to sitting and proving stuff from scratch, refusing to look it up in a textbook or ask for an answer until I had lost sleep over it for a couple of days. It works, believe me; what you get for free, doesn’t stay in the mind. That’s why I often give only a minimalistic hint and don’t give detailed, “spoon-feeding” solutions on the blog…a procedure I have no intention of changing.

regards
J

9. pragdeesh |

okay, i will try my best ðŸ™‚

10. pragdeesh |

AGNNNOO
N_ _ _ _ _ _
BEFORE THIS……………….. A G N N N O O—–60 WAYS
………………………………………G A N N N O O—–60 WAYS
NOW WE GET ———— N N N A G O O
BEFORE N O _ _ _ _ _ —————-N A G N N O O—30 WAYS
————————————————-N G A N N O O—30 WAYS
————————————————-N N N A G O O—60 WAYS
NOW WE GET N O O A G N N…….NEXT….N O A G N N O—12 WAYS
————————————————N O G A N N O—-12 WAYS
NOW WE GET..N O N N A G O……..NEXT…..N O N A G N O–0(since A is the first term)*3!-0 ways
———————————————————-N O N A G N O…..0* 2! WAYS(POSITION OF G)
N O N A G O N—-1*1 WAYS–1 WAY

did i got the concept?

11. pragdeesh |

sorry i made a mistake NNNAGOO IN THE FIRST AND IN MIDDLE IN MIDDLE IT WILL BE ………………………….. N N A G N O O—60 WAYS
.IN FIRST IT WILL BE LIKE ….N A G N N O O

12. himanshu |

hi,i am not an IMS student but still follows the concepts here on ur blog..i have one doubt reg question 2..u have mentioned that “words starting with E would be 7!=5040 as all the other 7 letters would be distinct” ..why we are not dividing it by 2!? there will still be 2Es? can u please help me with this?

• Because when we start with an E, there is only 1 E left in the remaining letters EGMORTY.

regards
J

13. himanshu |

ya i guess i have got it..i took the example of A, A and B and tried..when we start with A, the possible number of arrangements will be 2(AAB and ABA) and not 1..bt still if u think i have commited any mistake kindly help me..Thanks

14. himanshu |

why u have done 2*5! in question 2? not able to understand that ðŸ˜¦

• After G and E, the next letter we want is O. But before that we will have all words starting with GEE and GEM, each of which will give 5! words. Hence 2 * 5!

regards
J

15. Vaibhav |

In the second question should not the rank be 5306 itself as after T in the third last place RY is the first option?

• No, that 5306 is the number before the answer. Next one will be RY (note that the last term in the addition is 0 * 1!, to indicate that nothing more comes before RY).

regards
J

• Vaibhav |

Oh right. I misread it to be 0 factorial. Sorry for the inconvenience.