Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 4164 by Yozzii last updated on 30/Dec/15

Commented by Yozzii last updated on 30/Dec/15

Can anyone provide a well explained  proof of this formula? I don′t  understand why it works.

$${Can}\:{anyone}\:{provide}\:{a}\:{well}\:{explained} \\ $$$${proof}\:{of}\:{this}\:{formula}?\:{I}\:{don}'{t} \\ $$$${understand}\:{why}\:{it}\:{works}. \\ $$

Commented by prakash jain last updated on 30/Dec/15

I think it should be^(k−1) C_(n−1)  rather than ((k−1)/(n−1))

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:^{{k}−\mathrm{1}} {C}_{{n}−\mathrm{1}} \:\mathrm{rather}\:\mathrm{than}\:\frac{{k}−\mathrm{1}}{{n}−\mathrm{1}} \\ $$

Answered by prakash jain last updated on 31/Dec/15

★★★★....★  Say a total of k stars are k−balls.  To divide these into n−parts you need  n−1 bars to be placed in a total of k−1  position.  Number of ways to achieve^(k−1) C_(n−1)   Example 6 stars and 4 boxes.  ★★★★★★  ★∣★★∣★★∣★  (1,2,2,1)  ★∣★∣★∣★★★  (1,1,1,3)  To divide 6 stars among four boxes you  can 3 bars in 5 spaces between stars.  Since no box is empty max one bar can  be placed between 2 stars.

$$\bigstar\bigstar\bigstar\bigstar....\bigstar \\ $$$$\mathrm{Say}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:{k}\:\mathrm{stars}\:\mathrm{are}\:{k}−{balls}. \\ $$$$\mathrm{To}\:\mathrm{divide}\:\mathrm{these}\:\mathrm{into}\:{n}−\mathrm{parts}\:\mathrm{you}\:\mathrm{need} \\ $$$${n}−\mathrm{1}\:\mathrm{bars}\:\mathrm{to}\:\mathrm{be}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:{k}−\mathrm{1} \\ $$$$\mathrm{position}. \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{achieve}\:^{{k}−\mathrm{1}} {C}_{{n}−\mathrm{1}} \\ $$$$\mathrm{Example}\:\mathrm{6}\:\mathrm{stars}\:\mathrm{and}\:\mathrm{4}\:\mathrm{boxes}. \\ $$$$\bigstar\bigstar\bigstar\bigstar\bigstar\bigstar \\ $$$$\bigstar\mid\bigstar\bigstar\mid\bigstar\bigstar\mid\bigstar\:\:\left(\mathrm{1},\mathrm{2},\mathrm{2},\mathrm{1}\right) \\ $$$$\bigstar\mid\bigstar\mid\bigstar\mid\bigstar\bigstar\bigstar\:\:\left(\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{3}\right) \\ $$$$\mathrm{To}\:\mathrm{divide}\:\mathrm{6}\:\mathrm{stars}\:\mathrm{among}\:\mathrm{four}\:\mathrm{boxes}\:\mathrm{you} \\ $$$$\mathrm{can}\:\mathrm{3}\:\mathrm{bars}\:\mathrm{in}\:\mathrm{5}\:\mathrm{spaces}\:\mathrm{between}\:\mathrm{stars}. \\ $$$$\mathrm{Since}\:\mathrm{no}\:\mathrm{box}\:\mathrm{is}\:\mathrm{empty}\:\mathrm{max}\:\mathrm{one}\:\mathrm{bar}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{placed}\:\mathrm{between}\:\mathrm{2}\:\mathrm{stars}. \\ $$

Commented by Yozzii last updated on 30/Dec/15

Thanks a lot for the help!

$${Thanks}\:{a}\:{lot}\:{for}\:{the}\:{help}! \\ $$

Commented by Rasheed Soomro last updated on 31/Dec/15

EASY Explanation!

$$\mathcal{EASY}\:\mathcal{E}{xplanation}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com