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In how many ways can 10 identical tyres be distributed by the supplier

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fitzpratik
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In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   19 Feb 2019, 08:02
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In how many ways can 10 identical tyres be distributed by the supplier to 4 retail shops if any shop can get any number of tires?

A. \(\frac{13!}{10!3!}\)

B. \(\frac{10!}{7!3!}\)

C. \(\frac{10!}{7!4!}\)

D. \(\frac{11!}{7!4!}\)

E. \(\frac{13!}{10!4!}\)
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ScottTargetTestPrep
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   21 Feb 2019, 18:03
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fitzpratik wrote:
In how many ways can 10 identical tyres be distributed by the supplier to 4 retail shops if any shop can get any number of tires?

A. \(\frac{13!}{10!3!}\)

B. \(\frac{10!}{7!3!}\)

C. \(\frac{10!}{7!4!}\)

D. \(\frac{11!}{7!4!}\)

E. \(\frac{13!}{10!4!}\)


We can let T be a tire, so we have:

TTTTTTTTTT

Since any of the 4 shops can receive any number of tires (including 0), let’s use 3 strokes (|) to separate the tires. For example, we can have:

TTT|TTT|TT|TT or TTTTTTTTTT|||

That is, in the first example, the first two shops each receive 3 tires, and the last two shops each receive 2 tires. In the latter example, the first shop receives all 10 tires while the other 3 shops receive none.

Therefore, the question becomes how many ways can we arrange 10 Ts and 3 strokes. The answer is:

13!/(10!3!)

Answer: A
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   20 Feb 2019, 03:21
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fitzpratik wrote:
In how many ways can 10 identical tyres be distributed by the supplier to 4 retail shops if any shop can get any number of tires?

A. \(\frac{13!}{10!3!}\)

B. \(\frac{10!}{7!3!}\)

C. \(\frac{10!}{7!4!}\)

D. \(\frac{11!}{7!4!}\)

E. \(\frac{13!}{10!4!}\)


This is more of a formula based question. So better to memorize the below formula!

Number of ways of dividing 'n' identical objects into 'r' groups such that each group can contain any number of objects is given by

\(n+r-1_C_{r-1}\)

So,
The number of ways of dividing 10 identical tyres among 4 retail shops is

\(10+4-1_C_{4-1}\)

= \(13_{C_3}\)

= \(\frac{13!}{10!*3!}\)

Hence A is the correct answer.
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   21 Feb 2019, 23:37
chetan2u

Could you please give a clear explanation .
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   22 Feb 2019, 06:29
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fitzpratik wrote:
In how many ways can 10 identical tyres be distributed by the supplier to 4 retail shops if any shop can get any number of tires?

A. \(\frac{13!}{10!3!}\)

B. \(\frac{10!}{7!3!}\)

C. \(\frac{10!}{7!4!}\)

D. \(\frac{11!}{7!4!}\)

E. \(\frac{13!}{10!4!}\)


This is how you tackle distributing identical objects into distinct groups:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... inatorics-–-part-1/

Check out the method 2 of question 2. It explains the most direct method of handling questions of this type. You don't need to remember any formulae in that case.
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   24 Feb 2019, 03:51
ScottTargetTestPrep wrote:
fitzpratik wrote:
In how many ways can 10 identical tyres be distributed by the supplier to 4 retail shops if any shop can get any number of tires?

A. \(\frac{13!}{10!3!}\)

B. \(\frac{10!}{7!3!}\)

C. \(\frac{10!}{7!4!}\)

D. \(\frac{11!}{7!4!}\)

E. \(\frac{13!}{10!4!}\)


We can let T be a tire, so we have:

TTTTTTTTTT

Since any of the 4 shops can receive any number of tires (including 0), let’s use 3 strokes (|) to separate the tires. For example, we can have:

TTT|TTT|TT|TT or TTTTTTTTTT|||

That is, in the first example, the first two shops each receive 3 tires, and the last two shops each receive 2 tires. In the latter example, the first shop receives all 10 tires while the other 3 shops receive none.

Therefore, the question becomes how many ways can we arrange 10 Ts and 3 strokes. The answer is:

13!/(10!3!)

Answer: A



Hi TTP,

I find your method of this explanation very innovative !!!!
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   19 Sep 2020, 11:36
Asked: In how many ways can 10 identical tyres be distributed by the supplier to 4 retail shops if any shop can get any number of tires?

Number of ways = (10+4-1)C(4-1) = 13C3 = 13!/10!3!

IMO A
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink] New post   05 Apr 2024, 22:50
Using star and bars for unrestricted distribution:

10+4-1 C 4-1=13C3
A)
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Re: In how many ways can 10 identical tyres be distributed by the supplier [#permalink]
05 Apr 2024, 22:50
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