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02 Jun 2023, 08:38
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D
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Difficulty:
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Question Stats:
42% (02:58) correct
58%
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wrong
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Nancy is asked to select integers, between 0 and 10\(^1\)\(^3\). If she is instructed to select only those integers whose sum of digits is 3, then how many different integers can she select within the given range?
A. 235
B. 340
C. 377
D. 455
E. 533
A. 235
B. 340
C. 377
D. 455
E. 533
02 Jun 2023, 11:33
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ChandlerBong
Three possible cases in which the sum of the digits is three
1) The number consists of three 1s and all the remaining digits are 0s.
2) The number consists of one 2s, one 1s and all the remaining digits are 0s.
2) The number consists of one 3s and all the remaining digits are 0s.
Case 1: The number consists of three 1s and all the remaining digits are 0s
Min term = 1
Max terms = 9999..(upto 13 times)
The number of terms can be visualized as various arrangments of three 1s and ten 0s
Number of terms = \(\frac{13!}{ 3! * 10!}\)
= \(\frac{13 * 12 * 11 * 10!}{ 3! * 10!}\) = 286
Case 2: The number consists of one 2s, one 1s and all the remaining digits are 0s.
The number of terms can be visualized as various arrangments of one 2s, one 3s, and eleven 0s
Number of terms = \(\frac{13!}{ 3! * 10!}\)
= \(\frac{13 * 12 * 11!}{ 11!}\) = 156
Case 3: The number consists of one 3s and all the remaining digits are 0s.
The number of terms can be visualized as various arrangments of one 3s, and twelve 0s
Number of terms = \(\frac{13!}{ 3! * 10!}\)
= \(\frac{13 * 12!}{ 12!}\) = 13
Total number of terms = 286 + 156 +13 = 455
Option D
22 Jun 2023, 00:56
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gmatophobia
in case 2, why are we taking denominator as 3! * 10! instead of 2! * 11! and why does it say one 2s and one 3s, shouldn't it be one 1s?
