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23 Feb 2021, 04:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:26) correct
63%
(02:53)
wrong
based on 59
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If N denotes the number of different selections of 5 letters from the word MISSISSIPPI then N belongs to the set
(A) {15, 16, 17, 18, 19}
(B) {20, 21, 22, 23, 24}
(C) {25, 26, 27, 28, 29}
(D) {30, 31, 32, 33, 34}
(E) {35, 36, 37, 38, 39}
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) {15, 16, 17, 18, 19}
(B) {20, 21, 22, 23, 24}
(C) {25, 26, 27, 28, 29}
(D) {30, 31, 32, 33, 34}
(E) {35, 36, 37, 38, 39}
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Updated on: 22 Feb 2025, 05:01
Originally posted by GMATinsight on 23 Feb 2021, 04:14.
Last edited by Bunuel on 22 Feb 2025, 05:01, edited 2 times in total.
Last edited by Bunuel on 22 Feb 2025, 05:01, edited 2 times in total.
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Bunuel
Total letters in word MISSISSIPPI = 11
Frequency count is as follows
M - 1
I - 4
S - 4
P - 2
Case 1: 5 letter combination may have all 4 distinct letters with one letter repeated twice
Total such letter combination = 3 (two I, two S and two P)
Case 2: 5 letter combination may have all 3 distinct letters with two letters repeated twice
Total such letter combination = 3C2 (two letters who repeat out of 3 possible repeated letters)*2C1 (one of the two non repeated letter) = 6
Case 3: 5 letter combination may have all 3 distinct letters with one letter repeated thrice
Total such letter combination = 2C1 (letter which repeats thrice)*3C2 = 6
Case 4: 5 letter combination may have all 2 distinct letters with one letter repeated thrice and other repeated twice
Total such letter combination = 2 (letters which repeats thrice)*2(letters that repeat twice) = 4
Case 5: 5 letter combination may have all 2 distinct letters with one letter repeated four times and other repeated once
Total such letter combination = 2 (letters which repeats four times)*3 = 6
Total cases = 3+6+6+4+6 = 25
Answer: Option C
27 Feb 2021, 15:33
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Bunuel
There is 1 of M, 4 of I and S, and 2 of P. Note we are looking for selections of 5 letters, so order does not matter.
The first group of selections including M:
We need to fill in 4 spaces, we may have IIII, IIIS, IISS, ISSS, SSSS, IIIP, IISP, ISSP, SSSP, IIPP, ISPP, SSPP. 5 + 4 + 3 = 12 selections.
Next, we can repeat the pattern above but without M:
IIIIP through SSSSP is 5 selections.
IIIPP through SSSPP is 4 selections. Thus in total 9.
Finally I and S only has IIIII through SSSSS, which is 6 selections.
Therefore in total 12 + 9 + 6 = 27 selections.
Ans: C
