Permutation

Joy Shaha wrote:Q. In how many different ways can 4 MATH, 3 ENGLISH & 2 ANALYTICAL ABILITY books arranged in a row so that all books of the same branch are together?
A.864;
B.1484;
C.1726;
D.1728;
E.1734

Since all books of the same branch must be together, put the 3 different branches into 3 SEPARATE BLOCKS, as follows:
Let the math block = [ABCD]
Let the English block = [RST]
Let the analytical block = [XY].

Number of ways to arrange the 3 blocks [ABCD], [RST] and [XYZ] = 3!.
Within the math block, the number of ways to arrange books A, B, C and D = 4!.
Within the English block, the number of ways to arrange books R, S and T = 3!.
Within the analytical block, the number of ways to arrange books X and Y = 2!.
To combine the options above, we multiply:
3!4!3!2! = 6*24*6*2.
The product of the units digits = 6*4*6*2 = 24*12 = integer with a units digit of 8.
Thus, the correct answer choice must have a units digit of 8.

The correct answer is D.

Joy Shaha wrote:Q. In how many different ways can 4 MATH, 3 ENGLISH & 2 ANALYTICAL ABILITY books arranged in a row so that all books of the same branch are together?
A.864;
B.1484;
C.1726;
D.1728;
E.1734

(3!)^2 x 4! x 2 = 36 x 48 ==> some integer with a units digit of 8

Answer: D

Joy Shaha wrote:Q. In how many different ways can 4 MATH, 3 ENGLISH & 2 ANALYTICAL ABILITY books arranged in a row so that all books of the same branch are together?
A.864;
B.1484;
C.1726;
D.1728;
E.1734

Total books: 4 Math, 3 English, 2 Analytical Ability
Assume the books of each subject to be a bundle
Hence we have 3 bundles of Math, English and Analytical Ability

Total ways to arranging 3 bundles = 3!
Books inside each bundle can also be arranged.

Total ways = 3!*4!*3!*2! = 6*24*6*2 = 36*48 = 1728
By paying attention to the options, we can reach the correct one without calculating
As the answer would be a number rending with the digit 8 and there is only one such number in the given options.

Correct Option: D

Another approach:

Since you know the math book arrangements = 4 * 3 * 2 = 24, the answer must be divisible by 24. B, C, and E aren't, so you can throw those out.

We also know that since our total arrangements = 4 * 3 * 2 * 3 * 2 * 2 * 3 * 2, the answer must divide by 64. (4 * 2 * 2 * 2 * 2)

Only D does, so we're set.