Question

If 4 Maths books are selected from 6 different Maths different English books, how many ways can the seven om Maths books and 3 English books are chosen from 5 he seven books be arranged on a shelf: (a) If there are no restrictions? (4) • (3) - 76 = 756.000 (b) If the 4 Maths books remain together? (4) . (5.4!= 3400 (c) a Maths book is at the beginning of the shelf? (d) Maths and English books alternate. (9):13).46.3.- 216000 (e) A Maths is at the beginning and an English book is in the middle of the shelf

Answer 1

Answer:

Please see the answer below

Step-by-step explanation:

a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000

b. The number of ways such that the 4 math books remain together

The pattern is as follows:  MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM

Where M = Math’s Book and E= English Book.  

Number of ways = 4!*8!*4*150= 86400 ways.

c. The number of ways such that  math book is at the beginning of the shelf

The number of ways = 6!*4*150 = 432000

d. The number of ways such that  math and English books alternate

The number of ways = 150*4!*3! =2160 ways

e.  The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.