Number of ways of selecting 5 different letters = 5C5 = 1 way
Number of ways to select 2 similar and 3 different letter = 4C1* 4C3=16
Number of ways of selecting 2 similar + 2 more similar letter and 1 different letter = 4C2* 3C1= 18
Number of ways to select 3 similar and 2 different letter = 3C1* 4C2= 18
Number of ways to select 3 similar and another 2 other similar = 3C1*3C1=9
Number of ways to select 4 similar and 1 different letter = 2C1*4C1=8
ways of selecting 5 similar letters = 1
total ways = 1+16+18+18+9+8+1= 71
tough combination
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winniethepooh
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newgmattest
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Jim@Knewton
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Very interesting question Mariah!
Please see the image (solution structure) below to review along with the explanation.
The solution can be structured in 7 'selection conditions' (SC 1 to SC 7)
SC1: All five Ls (letters) are the same
Only possible for "A" = 1 letter => Number of ways = 1C1 = 1 way .............................(1)
SC2: Only 4 Ls are the same (Only possible with A and B) and one is different (1 out of 4 remaining, after A or B is selected)
=> Number of ways = 1*4C1 + 1*4C1 = 8 (Same as 2C1*4C1 = 8).............(2)
SC3: Only 3 Ls are the same (Possible with A, B and C) and two are diff.( 2 out of 4 remaining, after A or B or C is selected)
=> Number of ways = 1*4C2+1*4C2+1*4C2 = 18 (same as 3C1*4C2 = 18) .......(3)
SC4: 3 Ls are the same (Possible with A, B and C) and two others are same (Possible with A, B, C and D => 2 letters of 1 letter type from 3 remaining letter types)
=> Number of ways = 1*3C1+1*3C1+1*3C1 = 9 (Same as 3C1*3C1 = 9) .........(4)
SC5: 2 Ls are the same (Possible with A, B, C and D) and three are diff (same logic as above)
=> Number of ways = 1*4C3+1*4C3+1*4C3+1*4C3 = 16 (same as 4C1*4C3 = 16) .......(5)
SC6: 2 Ls are the same (Possible with A, B, C and D) and 2 other Ls are the same (Possible with A, B, C and D)and one different letter (same logic as above)
=> Number of ways = 1*3C1*3C1 +1*2C1*3C1 +1*1C1*3C1 = 18 (Same as 4C2*3C1= 18) .....(6)
[SC6 is a little tricky because we need to avoid repetition as shown above]
SC7: All 5 Ls are diff (Possible only with A, B, C, D and E)
=> Number of ways = 5C5 = 1 ...................................................(7)
Adding the number of ways from the 7 SCs = 1+8+18+9+16+18+1 = 71
Hope this helps...
Please see the image (solution structure) below to review along with the explanation.
The solution can be structured in 7 'selection conditions' (SC 1 to SC 7)
SC1: All five Ls (letters) are the same
Only possible for "A" = 1 letter => Number of ways = 1C1 = 1 way .............................(1)
SC2: Only 4 Ls are the same (Only possible with A and B) and one is different (1 out of 4 remaining, after A or B is selected)
=> Number of ways = 1*4C1 + 1*4C1 = 8 (Same as 2C1*4C1 = 8).............(2)
SC3: Only 3 Ls are the same (Possible with A, B and C) and two are diff.( 2 out of 4 remaining, after A or B or C is selected)
=> Number of ways = 1*4C2+1*4C2+1*4C2 = 18 (same as 3C1*4C2 = 18) .......(3)
SC4: 3 Ls are the same (Possible with A, B and C) and two others are same (Possible with A, B, C and D => 2 letters of 1 letter type from 3 remaining letter types)
=> Number of ways = 1*3C1+1*3C1+1*3C1 = 9 (Same as 3C1*3C1 = 9) .........(4)
SC5: 2 Ls are the same (Possible with A, B, C and D) and three are diff (same logic as above)
=> Number of ways = 1*4C3+1*4C3+1*4C3+1*4C3 = 16 (same as 4C1*4C3 = 16) .......(5)
SC6: 2 Ls are the same (Possible with A, B, C and D) and 2 other Ls are the same (Possible with A, B, C and D)and one different letter (same logic as above)
=> Number of ways = 1*3C1*3C1 +1*2C1*3C1 +1*1C1*3C1 = 18 (Same as 4C2*3C1= 18) .....(6)
[SC6 is a little tricky because we need to avoid repetition as shown above]
SC7: All 5 Ls are diff (Possible only with A, B, C, D and E)
=> Number of ways = 5C5 = 1 ...................................................(7)
Adding the number of ways from the 7 SCs = 1+8+18+9+16+18+1 = 71
Hope this helps...
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Jim@Knewton
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Yes, this is a tough question, IMO definitely in the 700+ category. The question here is structured to require us to compute at different levels and integrate. There doesn't appear to be any (easier) way to avoid the probability computations.
Practice will help improve timing - review other probability questions solved on this board /others and practice, practice, practice!
Practice will help improve timing - review other probability questions solved on this board /others and practice, practice, practice!
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gmatboost
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I am willing to go out on a limb and say that you are NOT going to see this question on the GMAT. Beware of overly complex practice permutations/combinations questions. The question designer is often looking to create a "hard" practice question, and it's pretty easy to cross the line from "hard" to "not realistic."
Perhaps I'm wrong, but I'd have to see some examples of actual test questions that require summing anything close to SEVEN different combinations to feel like this is not the case.
Perhaps I'm wrong, but I'd have to see some examples of actual test questions that require summing anything close to SEVEN different combinations to feel like this is not the case.
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GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.
Also, check out the most useful GMAT Math blog on the internet here.
