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# CAT 2021 Quant Question [Slot 1] with Solution 07

Question
The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is
Option: 47
Solution:

The possible arrangements are of the form
35 _ Can be chosen in 6 ways.
35 _ _ We can choose 2 out of the remaining 6 in $^6{C_2} = 15$ways. We remove 1 case where 7 and 8 are together to get 14 ways.
35 _ _ _We can choose 3 out of the remaining 6 in $^6{C_3} = 20$ways. We remove 4 cases where 7 and 8 are together to get 16 ways.
35 _ _ _ _We can choose 4 out of the remaining 6 in $^6{C_4} = 15$ways. We remove 6 case where 7 and 8 are together to get 9 ways.
35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.
Thus, total number of cases = 6+14+16+9+2 = 47.
Alternatively,
The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.
Considering the cases, only 7 is selected.
We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.
Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have
Each of the numbers can either be selected or not selected and we have 4 numbers :
Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.
SImilarly, including only 8, we have 16 more possibilities.
Cases including neither 7 nor 8.
We must have 3 and 5 in the group but there must be no 7 and 8 in the group.
Hence we have 3 5 _ _ _ _.
For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.
= 16 possibilities
But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.
Hence 16 - 1 = 15 cases.
Total = 16+15+16 = 47 possibilities
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