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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 18 Jun 2018, 06:37
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At a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmatian division. If the top 3 dogs in each division receive first-, second-, and third-place ribbons respectively, with no other dogs receiving a prize, how many different ways can ribbons be awarded to winners in the two divisions together?


A. 28,800

B. 14,400

C. 720

D. 400

E. 36
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B
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 18 Jun 2018, 07:13
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Solution



Given:
    • 6 dogs entered in beagle division
    • 6 dogs entered in dalmatian division
    • Top 3 dogs in each division receive 1st, 2nd and 3rd place ribbons respectively

To find:
    • In how many different ways the ribbons can be awarded to winners in two divisions together

Approach and Working:
In the beagle division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\) = 120

Similarly, in the Dalmatian division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\)= 120

    • Hence, the total number of different ways ribbons can be awarded to the winners, in both the divisions together = 120 * 120 = 14400

Hence, the correct answer is option B.

Answer: B
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 18 Jun 2018, 20:09
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Bunuel
At a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmatian division. If the top 3 dogs in each division receive first-, second-, and third-place ribbons respectively, with no other dogs receiving a prize, how many different ways can ribbons be awarded to winners in the two divisions together?


A. 28,800

B. 14,400

C. 720

D. 400

E. 36
Show more
Fundamental Counting Principle

In the beagle division, 6 dogs vie for 1st, 2nd, and 3rd Place (For 1st place, there are 6 choices; for 2nd place, there are 5 choices; for 3rd place there are 4)


__6__ * __5__ * __4__ = 120

In the Dalmatian division, the situation is identical

__6__ * __5__ * __4__ = 120

Total number of ways that ribbons can be awarded to winners in the two divisions together: 120*120 = 14,400

Answer B
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 20 Jun 2018, 17:00
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Bunuel
At a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmatian division. If the top 3 dogs in each division receive first-, second-, and third-place ribbons respectively, with no other dogs receiving a prize, how many different ways can ribbons be awarded to winners in the two divisions together?


A. 28,800

B. 14,400

C. 720

D. 400

E. 36
Show more

Since the order of finishing in each division is important, we use permutations. The number of ways in which the awards can be provided is 6P3 x 6P3 = 6!/3! x 6!/3! = 6 x 5 x 4 x 6 x 5 x 4 = 14,400.

Answer: B
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 14 Jul 2019, 15:05
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Solution



Given:
    • 6 dogs entered in beagle division
    • 6 dogs entered in dalmatian division
    • Top 3 dogs in each division receive 1st, 2nd and 3rd place ribbons respectively

To find:
    • In how many different ways the ribbons can be awarded to winners in two divisions together

Approach and Working:
In the beagle division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\) = 120

Similarly, in the Dalmatian division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\)= 120

    • Hence, the total number of different ways ribbons can be awarded to the winners, in both the divisions together = 120 * 120 = 14400

Hence, the correct answer is option B.

Answer: B
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Hello EgmatQuantExpert!

Why should we multiply by 3! after 6C3?

Kind regards!
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 14 Jul 2019, 19:13
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Solution



Given:
    • 6 dogs entered in beagle division
    • 6 dogs entered in dalmatian division
    • Top 3 dogs in each division receive 1st, 2nd and 3rd place ribbons respectively

To find:
    • In how many different ways the ribbons can be awarded to winners in two divisions together

Approach and Working:
In the beagle division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\) = 120

Similarly, in the Dalmatian division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\)= 120

    • Hence, the total number of different ways ribbons can be awarded to the winners, in both the divisions together = 120 * 120 = 14400

Hence, the correct answer is option B.

Answer: B

Hello EgmatQuantExpert!

Why should we multiply by 3! after 6C3?

Kind regards!
Show more

jfranciscocuencag : it is because the order is important here. If abc and acb are the same, then we use combination i.e. the order is not important. But if abc and acb are different that means the order is important. SO We need in how many ways "a"."b" and "c" can be listed.
n this question if D1, D2 and D3 and D1, D3 and D2 are different. So 3! is multiplied.

EgmatQuantExpert / chetan2u: My doubt is why are we multiplying 120*120. I thought they are independent events because the award is happening for 2 other divisions. So I expected 2*120 would be the right answer. Could you please explain what I am missing.
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 21 Jul 2019, 01:59
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Solution



Given:
    • 6 dogs entered in beagle division
    • 6 dogs entered in dalmatian division
    • Top 3 dogs in each division receive 1st, 2nd and 3rd place ribbons respectively

To find:
    • In how many different ways the ribbons can be awarded to winners in two divisions together

Approach and Working:
In the beagle division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\) = 120

Similarly, in the Dalmatian division, we have 6 dogs
    • Therefore, the number of ways 3 of them can be awarded ribbons = \(^6C_3 * 3! = ^6P_3\)= 120

    • Hence, the total number of different ways ribbons can be awarded to the winners, in both the divisions together = 120 * 120 = 14400

Hence, the correct answer is option B.

Answer: B

Hello EgmatQuantExpert!

Why should we multiply by 3! after 6C3?

Kind regards!

jfranciscocuencag : it is because the order is important here. If abc and acb are the same, then we use combination i.e. the order is not important. But if abc and acb are different that means the order is important. SO We need in how many ways "a"."b" and "c" can be listed.
n this question if D1, D2 and D3 and D1, D3 and D2 are different. So 3! is multiplied.

EgmatQuantExpert / chetan2u: My doubt is why are we multiplying 120*120. I thought they are independent events because the award is happening for 2 other divisions. So I expected 2*120 would be the right answer. Could you please explain what I am missing.
Show more

chetan2u, Gladiator59, VeritasKarishma, Bunuel, generis --

Could you please help me to understand the doubt I have?

Regards,
Arup Sarkar
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 21 Jul 2019, 22:00
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Bunuel
At a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmatian division. If the top 3 dogs in each division receive first-, second-, and third-place ribbons respectively, with no other dogs receiving a prize, how many different ways can ribbons be awarded to winners in the two divisions together?


A. 28,800

B. 14,400

C. 720

D. 400

E. 36
Show more

Say there are 6 Beagles - B1, B2, B3, B4, B5 and B6
And there are 6 Dalmatians - D1, D2, D3, D4, D5 and D6

Question: How many different ways can ribbons be awarded to winners in the two divisions together?
What constitutes ribbons awarded in different ways in the two divisions together? BRank1, BRank2, BRank3, DRank1, DRank2, DRank3

- B2, B3, B5, D3, D6, D1
- B3, B2, B5, D6, D3, D1 (another way - First and second ranks of Beagles are interchanged and first and second ranks of Dalmatians are interchanged)
- B3, B2, B5, D6, D3, D5 (another way - Note that just changing the 3rd rank of Dalmatians changes the way in which ribbons are distributed in the two divisions together)

Of the 6 Beagles, select 3 and arrange them in 3! ways in first 3 spots - 6C3 * 3!
Of the 6 Dalmatians, select 3 and arrange them in 3! ways in the last 3 spots - 6C3 * 3!

Total ways = 6C3 * 3! * 6C3 * 3! (the two will be multiplied because the result includes both simultaneously)

Answer (B)
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At a certain pet show, exactly 6 dogs are entered in the beagle divisi 16 Apr 2026, 05:10
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