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In how many ways can the letters in the name BELLA be arrang
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04 Apr 2019, 09:31
04 Apr 2019, 09:31
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In how many ways can the letters in the name BELLA be arranged?
A. 5
B. 15
C. 30
D. 60
E. 120
A. 5
B. 15
C. 30
D. 60
E. 120
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Joined: 21 Nov 2018
Posts: 24
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GRE 1: Q170 V170
Re: In how many ways can the letters in the name BELLA be arrang
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04 Apr 2019, 11:30
04 Apr 2019, 11:30
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We need to add to this question the fact that the two "L" letters are identical. In this case what we have is a permutation with identical elements. This occurs when you have two or more elements of a list that you cannot tell apart and therefore do not generate new orders or arrangements (permutations).
You can see that this is an identical elements problem in the following example. If we have the letters arranged ELLBA. I will now switch just the order of the two Ls. ELLBA. Can you tell the difference? No? Well that makes this a permutation with identical elements.
Here is the formula. N! / A! B! C! N = the number of items pool that you draw from - in this case there are 5 letters. So N = 5. A, B, C etc. are the numbers of identical elements with each taken as a separate factorial. In this case we just have 2 identical Ls so the denominator is just 2!.
So we have 5! / 2! .... 5! = 5 * 4 * 3 *2 *1 = 120 and 2! = 2*1 = 2. So 120/ 2 = 60. The answer is D.
What if we apply the same identical elements formula to MISSISSIPPI?
Now we have N! = 11! for the total of 11 letters.
We have A! = 4! for 4 identical Ss. And B! = 4! for 4 identical Is and C! = 2! for 2 identical Ps.
So this becomes 11! / 4! 4! 2!. Before you calculate you can reduce. So 11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4!. Cancel that 4! with one of the 4! in the denominator. Now you have 11 * 10 * 9 * 8 * 7 * 6 * 5 / 4 * 3 * 2 *1 * 2 * 1......3 * 2 = 6 so cancel that out from the numerator. 4 * 2 = 8 so cancel the 8 as well. You are now left with 11 * 10 * 9 * 7 * 5. Which equals = 34, 650.
You can see that this is an identical elements problem in the following example. If we have the letters arranged ELLBA. I will now switch just the order of the two Ls. ELLBA. Can you tell the difference? No? Well that makes this a permutation with identical elements.
Here is the formula. N! / A! B! C! N = the number of items pool that you draw from - in this case there are 5 letters. So N = 5. A, B, C etc. are the numbers of identical elements with each taken as a separate factorial. In this case we just have 2 identical Ls so the denominator is just 2!.
So we have 5! / 2! .... 5! = 5 * 4 * 3 *2 *1 = 120 and 2! = 2*1 = 2. So 120/ 2 = 60. The answer is D.
What if we apply the same identical elements formula to MISSISSIPPI?
Now we have N! = 11! for the total of 11 letters.
We have A! = 4! for 4 identical Ss. And B! = 4! for 4 identical Is and C! = 2! for 2 identical Ps.
So this becomes 11! / 4! 4! 2!. Before you calculate you can reduce. So 11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4!. Cancel that 4! with one of the 4! in the denominator. Now you have 11 * 10 * 9 * 8 * 7 * 6 * 5 / 4 * 3 * 2 *1 * 2 * 1......3 * 2 = 6 so cancel that out from the numerator. 4 * 2 = 8 so cancel the 8 as well. You are now left with 11 * 10 * 9 * 7 * 5. Which equals = 34, 650.
Re: In how many ways can the letters in the name BELLA be arrang
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04 Apr 2019, 11:54
04 Apr 2019, 11:54
Expert Reply
Tricky when the letters are repeated.
Awesome.
Awesome.
