Math, asked by adduarru321, 10 months ago

A sleeper coach in a train has 8 lower berths, and 10 upper berths. Find the number of ways of arranging 18 persons in it, if 3 children want to go upper berth and 4 old people cannot go to the upper berth. A sleeper coach in a train has 8 lower berths, and 10 upper berths. Find the number of ways of arranging 18 persons in it, if 3 children want to go upper berth and 4 old people cannot go to the upper berth.

Answers

Answered by amitnrw
1

¹¹C₄ * 10!  * 8 !  is the number of ways   arranging 18 persons in 8 lower berths, and 10 upper berths when 3 children want to go upper berth & 4 old people cannot go to the upper berth.

Step-by-step explanation:

3 children want to go upper berth

4 Old People can not go to upper berth means will go to Lower berth

Remaining people = 11

out of 11   , 4 need to go lower berth  &  7 to upper berth

this Selection can be done by

¹¹C₄ or ¹¹C₇   Ways

8 Lower berths can be arranged in 8! ways

10 upper berths can be arranged in 10! Ways

Total arrangement = ¹¹C₄ * 10!  * 8 !

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Answered by dk6060805
1

4.828336e13 is the Answer

Step-by-step explanation:

As per the Given conditions,

I Condition says- 3 Children = Upper Births

II Condition says 4 Old People = Lower Births

III Condition says, Off 11 people,

4 People = Lower Births = ^{11}C_4

7 People = upper  Births = ^{11}C_7

Further, 10 Upper births can be arranged in 10! ways

& 8 Lower Births can be arranged in 8! ways.

Hence, Total Arrangements = ^{11}C_4 \times 10!\times 8!

                                                = (11! x 10! x 8! )/ (11 - 4)!

                                                = 11! x 10! x 8! / 7! x 4!

                                                = (11 x 10 x 9 x 8 ) x (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) x (8 x 7 x 6 x 5)

                                                = 7920 x 3628800 x 1680

                                                = 4.828336e13

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