advance math mcq pdf Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Answer: (b)

Number of ways of selecting 5 guests from nine friends = $^9C_5$

Out of these, $^7C_3$ ways are those in which two of the friends occur together [3 more persons to be selected out of remaining 7]

∴ Number of ways, in which two of the friends will not attend the party together = $^9C_5 – ^7C_3$ = 91.

Answer: (b)

These can be two cases :

Case I - One box contain 3 balls and rest two Contains 1 ball each.

Case II - One box contain 1 ball and rest two Contains 2 balls each.

Case 1 : Number of ways = $^5C_3 × ^2C_1 × ^1C_1$ = 20

Now, these 3 boxes can be arranged in ${3!}/{2!}$ among themselves, as two of them contains similar number of balls.

So, number of ways = 20 × ${3!}/{2!}$ = 60

Case II : Number of ways = $^5C_1 × ^4C_2 × ^2C_2$ = 30

Also, these 3 boxes can be arranged among themselves in ${3!}/{2!}$, as two of them contains similar number of balls.

Thus, number of ways = 30 × ${3!}/{2!}$ = 90

Now, either of case (I) or case (II) is possible,

Hence, total number of ways = 60 + 90 = 150

Answer: (a)

Letters of the word PARKAR written in alphabetical

order are A A K P R R

Number of words starting with A is = 60

Number of words starting with K is = 30

Number of words starting with PAA is = 3

Number of words starting with PAK is = 3

Number of words starting with PARA is = 2

Number of words starting with PARKAR is = 1

∴ Rank of word PARKAR is 99

Answer: (b)

X - X - X - X - X. The four digits 3, 3, 5,5 can be arranged

at (-) places in ${4!}/{2!2!}$ = 6 ways.

The five digits 2, 2, 8, 8, 8 can be arranged at (X) places in ${5!}/{2!3!}$ ways = 10 ways

Total no. of arrangements = 6 × 10 = 60 ways

Answer: (c)

In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.

Thus, (7/15) × (8/15) = 56/225.