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Two girls have picked $10$ roses, $15$ sunflowers and $15$ daffodils. What is the number of ways they can divide the flowers among themselves?

1. $1638$
2. $2100$
3. $2640$
4. None of the above
edited | 3.6k views
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In Gate question it's 14 daffodils not 15. After taking 14 daffodils answer is coming (C).
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Division and Distribution of Identical Objects
Case 1

Number of ways in which n identical things can be divided into r groups, if blank groups are allowed (here groups are numbered, i.e., distinct)

= Number of ways in which n identical things can be distributed among r persons, each one of them can receive 0,1,2 or more items

= (n+r-1)C(r-1)

can we do like this here girls =2  lets  these two girls are two groups   ,  10 roses ,
so   (10 + 2 -1) C(2-1)  =  11
and so on
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There is no clarity in the question, I guess.
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why are you considering flowers as identical objects??

For each flower type, say there are $n$ number of flowers. We apply star and bars method for each flower type. $n$ flowers of a type will generate $(n+1)$ spaces we just need to place one bar which will separate them into $2$ for the two girls. To do that we need to select a position:

For roses: $\binom{10+1}{1}$
For sunflowers: $\binom{15+1}{1}$
For daffodils: $\binom{15+1}{1}$

Total number of ways distribution can take place $= 11 \times 16 \times 16 = 2816.$

Correct Answer: $D$
edited
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pretty simplified.

Number of ways roses can be distributed $= \{ (0, 10), (1, 9), (2, 8), \ldots,(10, 0) \}$ - $11 \text{ ways}$
Similarly, sunflowers and daffodils can be distributed in $16$ ways each
So, total number of ways $= 11 \times 16 \times 16 = 2816.$
edited by
Combination with repeated objects rule: N objects and N objects can repeat then total number of r combination = (n+r-1)Cr

Here there are 2 girls. So, n =2

1. 10 Roses

So r=10. Substitute in the formula, we get, (2+10-1)C10=11  -------- 1

2. 15 Sunflowers

So, r=15. So, (2+15-1)C15=16 --------------- 2

3. 15 Daf

So, r=15. So, (2+15-1)C15=16 ------------- 3

Total = 1*2*3=11*16*16=2816.

Therefore, Option D.

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