2

Complete Question:

There are $m$ bags which are numbered by $m$ consecutive positive integers starting with the number $k$. Each bag contains as many different flowers as the number labeled against the bag. A boy has to pick up k flowers from any one of the bags. In how many different ways he can do the work?

My Try

The bags are numbered $k,k+1,k+2,k+3,...,k+(m-1)$

This gives the total number of flowers as $${k+(k+1)+(k+2)+...+(k+(m-1))}=mk+ \dfrac{m(m-1)}{2}$$

I am having confusion in calculating the total number of objects and no. of objects to be chosen out of those.

I have tried by choosing the bag in $^mC_1$ ways and total ways of choosing $k$ objects out of those is by $${^kC_k\ +\ ^{k+1}C_k\ +\ ^{k+2}C_k\ +\cdots +\ ^{k+(m-1)}C_k}\ \text{ways}$$

which gives the complete ways as $$^mC_1\ \times\ \{^kC_k\ +\ ^{k+1}C_k\ +\ ^{k+2}C_k\ +\cdots +\ ^{k+(m-1)}C_k\}$$

I am not confident if the solution upto this point is correct and if it is then how can I simplify it further?

N. F. Taussig
  • 76,571
Harshit Bhardwaj
  • 115
  • As bags are numbered, they are distinct. It also mentions different flowers so I am assuming all flowers in each bag are distinct too. Can you confirm? – Math Lover Aug 14 '20 at 16:52
  • Yes the flowers must be distinct too if its phrased that way. – Harshit Bhardwaj Aug 14 '20 at 16:57
  • 2
    If that is the case, there is only one way to choose one specific numbered bag so your addition will do. You should not multiply by $mC1$. – Math Lover Aug 14 '20 at 16:57
  • 1
    take another example - where all $m$ bags contained $k$ flowers. What will be your answer? $^mC_1.^kC_k$ or ($^kC_k+^kC_k+..+^kC_k$ $m$ times. I hope that clarifies. – Math Lover Aug 14 '20 at 17:00
  • @Math Lover You mean that the ways of choosing the bag won't matter in making the selections. Let me rephrase that as "If the boy is selecting the flowers out of the bag it means that he has already selected a bag in 1 way only as all bags have at least k flowers and he doesn't need any other way to select a bag." Also don't $^mC_1. ^kC_k$ and $^kC_k + ^kC_k + \cdot +^kC_k m$ times imply the same thing? That is like first we select a bag in $^mC_1$ way and out of that we select $k$ flowers in $^kC_k$ ways and we multiply those for total ways or keep selecting $k$ flower out of $k$ flowers cont – Harshit Bhardwaj Aug 14 '20 at 17:28
  • 1
    Does it mean that my answer is only $^kC_k\ +\ ^{k+1}C_k\ +\ \cdots\ +\ ^{k+(m-1)}C_k\ ?$ How could I simplify it further to get only a single term of $^nC_r$ form? – Harshit Bhardwaj Aug 14 '20 at 17:41

2 Answers2

2

Yes your answer will be without multiplying by $^mC_1$. That is what I tried to convey by the other example in my comments. To your other question on simplifying the sum -

$\sum_{i=0}^{m-1}\binom{k+i}{k} = \binom{m+k}{k+1}$

This is known as the hockey-stick identity. Please see https://en.wikipedia.org/wiki/Hockey-stick_identity.

The proof of it can also be found on MSE - Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$.

Math Lover
  • 51,819
0

We do not need to multiply by $^mC_1$.

We solve this as:

selecting $k$ flowers out of each bag are $^kC_k\ +\ ^{k+1}C_k\ +\ \cdots\ +\ ^{k+(m-1)}C_k$ ways

The answer is $$^kC_k\ +\ ^{k+1}C_k\ +\ \cdots\ +\ ^{k+(m-1)}C_k$$

To Simplify: we know that $^nC_n\ +\ ^{n+1}C_n\ +\ \cdots\ +\ ^{n+k}C_n\ =\ ^{n+k+1}C_{n+1}$

Hence, our answer is $$^{k+(m-1)+1}C_{k+1}=\ ^{k+m}C_{k+1}$$

Also thanks to @Math Lover

Harshit Bhardwaj
  • 115