# CAT 2021 Question Paper | Quant Slot 3

###### CAT Previous Year Paper | CAT Quant Questions | Question 4

CAT 2021 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 4 : A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

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We will select the 4 digits first and arrange them later.
Out of the 4 digits, one of them should be 2 and one of them should be 3.
2, 3, , .
So, we just need to select the other two digits…
The two digits could be (1,1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3).
So, the selection of numbers could be…
2, 3, 1, 1
2, 3, 2, 2
2, 3, 3, 3
2, 3, 1, 2
2, 3, 1, 3
2, 3, 2, 3
Each of these selections could be re-arranged in a number of ways.
2, 3, 1, 1 ($$frac { 4 ! } { 2 ! }$ = 12 ways) 2, 3, 2, 2$$$frac { 4 ! } { 3 ! }$ = 4 ways) 2, 3, 3, 3$$$frac { 4 ! } { 3 ! }$ = 4 ways) 2, 3, 1, 2$$$frac { 4 ! } { 2 ! }$ = 12 ways) 2, 3, 1, 3$$$frac { 4 ! } { 2 ! }$ = 12 ways) 2, 3, 2, 3$$$frac { 4 ! } { 2 ! * 2 ! }$ = 6 ways) So total number of possibilities =$12 + 4 + 4 + 12 + 12 + 6) = 50 ways.
Alternate method:
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
Think why we need to add (Arrangements with only 1)!
_, _, _, _
(All possible arrangements) = 34
Each blank could be any one of 1, 2 or 3.
(Arrangements with either 1 or 2) = 24
Each blank could be any one of 1 or 2.
(Arrangements with either 1 or 3) = 24
Each blank could be any one of 1 or 3.
(Arrangements with only 1) = 1
Each blank is filled with 1.
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
(Arrangements with at least one 2 and one 3) = 34 - 24 - 24 + 1
(Arrangements with at least one 2 and one 3) = 81 - 16 - 16 + 1
(Arrangements with at least one 2 and one 3) = 82 - 32 = 50

The question is " A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is "

##### Hence, the answer is '50'

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