18 thoughts on “PnC Examples 4

  1. 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

  3. 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??

    • 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

  4. 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 ?

  5. 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 😛

      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

  6. 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?

  7. 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

  8. 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?

  9. 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

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