Homework on Pearson

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Mathematics

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I need help on these assignments I am a bit confused. Can you provide a word document on how to work the problem along with the question. So I can use this on my final exam. These assignments are not due until 4/17/2019 @11:59pm.

Complete sections

1. Section 14.2

2. Section 14.3

3. Section 14.4

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Explanation & Answer

Section 14.2, 14.3 and 14.4 are completed.😊

Sol: The sample space of two dice is given

Let E be the event for a score of at least 5. Let n(E) total number of outcomes resulting a
score of at least 5 and n(S) is the total number of outcomes.
𝑛(𝐸) = 30
𝑛(𝑆) = 36
So,
𝑃(𝐸) =

Hence.

𝑛(𝐸)
𝑛(𝑆)

𝑃(𝐸) =

30
36

𝑃(𝐸) =

5
6

Sol: Let n(A) be the total number of students speaking English, n(B) be the total number of
students speaking Japanese, n(A or B) be the total number of students speaking either
English or Japanese and n(A and B) be the total number of students that speak both
languages. So,
𝑛(𝐴) = 20
𝑛(𝐵) = 14
𝑛(𝐴 𝑜𝑟 𝐵) = 28
𝑛(𝐴 𝑎𝑛𝑑 𝐵) =?
(a)
As we know that,
𝑛(𝐴 𝑜𝑟 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 𝑎𝑛𝑑 𝐵)
So,
𝑛(𝐴 𝑎𝑛𝑑 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 𝑜𝑟 𝐵)
𝑛(𝐴 𝑎𝑛𝑑 𝐵) = 20 + 14 − 28
𝑛(𝐴 𝑎𝑛𝑑 𝐵) = 6
(b)

From the diagram, we can see that,

𝑛(𝐴) = 20
𝑛(𝐴 𝑎𝑛𝑑 𝐵) = 6
we can find the students who only speak English not Japanese, by subtracting n(A and B)
from n(A). So,
𝑂𝑛𝑙𝑦 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑠𝑝𝑒𝑎𝑘𝑖𝑛𝑔 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 = 𝑛(𝐴) − 𝑛(𝐴 𝑎𝑛𝑑 𝐵)
𝑂𝑛𝑙𝑦 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑠𝑝𝑒𝑎𝑘𝑖𝑛𝑔 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 = 20 − 6
𝑂𝑛𝑙𝑦 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑠𝑝𝑒𝑎𝑘𝑖𝑛𝑔 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 = 14
Hence.

Sol: Total number of hearts are 13 and total number of face cards are 12 in the deck of 52 playing
card. Let n(A) be the total number of heart card, n(B) be the total number of face cards, n(A
and B) be the total number of cards which are both heart and face card and n(A or B) be the
total number of cards which is either heart or face card. So,
𝑛(𝐴) = 13
𝑛(𝐵) = 12
𝑛(𝐴 𝑎𝑛𝑑 𝐵) = 3
𝑛(𝐴 𝑜𝑟 𝐵) =?
As we know that
𝑛(𝐴 𝑜𝑟 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 𝑎𝑛𝑑 𝐵)
𝑛(𝐴 𝑜𝑟 𝐵) = 13 + 12 − 3
𝑛(𝐴 𝑜𝑟 𝐵) = 22
Hence.

Sol: Let E1 be the event for a card to be black king and E2 be the event for a card to be black ace.

Then n(E1) be the total number of black king cards and n(E2) be the total number of black
ace card. n(S) be the total number of cards. So,
𝑛(𝐸1 ) = 2
𝑛(𝐸2 ) = 2
𝑃(𝐸1 𝑜𝑟 𝐸2 ) = 𝑃(𝐸1 ) + 𝑃(𝐸2 )
𝑃(𝐸1 𝑜𝑟 𝐸2 ) =

𝑛(𝐸1 ) 𝑛(𝐸2 )
+
𝑛(𝑆)
𝑛(𝑆)

𝑃(𝐸1 𝑜𝑟 𝐸2 ) =

2
2
+
52 52

𝑃(𝐸1 𝑜𝑟 𝐸2 ) =

2+2
52

𝑃(𝐸1 𝑜𝑟 𝐸2 ) =

4
52

𝑃(𝐸1 𝑜𝑟 𝐸2 ) =

1
13

Hence.

Sol: (a)
As repetition is not allowed, then first digit will be chosen from given 7 numbers, second
digit will be chosen from remaining 6 numbers, third digit will be chosen from remaining 5
numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 7 × 6 × 5 = 210
(b)

If the first digit is 2, then the second digit will be chosen from remaining 6 numbers and
third digit will be chosen from remaining 5 numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 2 = 6 × 5 = 30
If the first digit is 3, then the second digit will be chosen from remaining 6 numbers and
third digit will be chosen from remaining 5 numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 3 = 6 × 5 = 30
If the first digit is 4, then the second digit will be chosen from remaining 6 numbers and
third digit will be chosen from remaining 5 numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 4 = 6 × 5 = 30
So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 2,3 𝑜𝑟 4 = 30 + 30 + 30
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 2,3 𝑜𝑟 4 = 90
(c)
The three-digit number will be even if it ends with 2, 4 or 6.
If the number ends with 2, then the first digit will be chosen from remaining 6 numbers and
second digit will be chosen from remaining 5 numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑒𝑛𝑑𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 2 = 6 × 5 = 30
If the number ends with 4, then the first digit will be chosen from remaining 6 numbers and
second digit will be chosen from remaining 5 numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑒𝑛𝑑𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 4 = 6 × 5 = 30
If the number ends with 6, then the first digit will be chosen from remaining 6 numbers and
second digit will be chosen from remaining 5 numbers. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑒𝑛𝑑𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 6 = 6 × 5 = 30
So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑣𝑒𝑛 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 = 30 + 30 + 30
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑣𝑒𝑛 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 = 90
Hence.

Sol: (a)
If repetition is allowed, then all the letters of 5-letter code will be chosen from 24
alphabets. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑜𝑑𝑒 = 24 × 24 × 24 × 24 × 24
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑜𝑑𝑒 = 245
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑜𝑑𝑒 = 7962624
(b)
If repetition is not allowed, then first letter will be chosen from 24 alphabets, second letter
will be chosen from remaining 23 alphabets, third letter will be chosen from remaining 22
alphabets, fourth letter will be chosen from remaining 21 alphabets and fifth letter will be
chosen from remaining 20 alphabets. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑜𝑑𝑒 = 24 × 23 × 22 × 21 × 20
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑙𝑒𝑡𝑡𝑒𝑟 𝑐𝑜𝑑𝑒 = 5100480
Hence.

Sol: (a)
If repetition is allowed, then all of the first four digits will be chosen from the 5 odd digits
which are 1, 3, 5, 7, 9 and all of the last two digits will be chosen from the 5 even digits which
are 0, 2, 4, 6, 8. So,

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑥 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 5 × 5 × 5 × 5 × 5 × 5
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑥 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 56
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑥 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 15625
(b)
If the repetition is not allowed, then the first digit will be chosen from 5 odd digits, second
digit will be chosen from remaining 4 odd digits, third digit will be chosen from remaining 3
odd digits, fourth digit will be chosen from remaining 2 odd digits, fifth digit will be chosen
from five even digits and sixth digit will be chosen from remaining 4 even digits. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑥 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 5 × 4 × 3 × 2 × 5 × 4
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑥 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 2400
Hence.

Sol: The numbers that appear on a die are 1, 2, 3, 4, 5, 6.
(a)
The number that will appear on first die will be from 6 numbers above. Because the
numbers on all dice should be different, so, the number that will appear on second die will
be from remaining 5 of the numbers mentioned above. And the number that will appear on
third dice will be from remaining 4 numbers mentioned above. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 3 𝑑𝑖𝑐𝑒 𝑡𝑜 ℎ𝑎𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 6 × 5 × 4
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 3 𝑑𝑖𝑐𝑒 𝑡𝑜 ℎ𝑎𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 120
(b)
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑒𝑎𝑟 𝑜𝑛 𝑓𝑜𝑢𝑟 𝑑𝑖𝑐𝑒 = 6 × 6 × 6 × 6
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑒𝑎𝑟 𝑜𝑛 𝑓𝑜𝑢𝑟 𝑑𝑖𝑐𝑒 = 64
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑒𝑎𝑟 𝑜𝑛 𝑓𝑜𝑢𝑟 𝑑𝑖𝑐𝑒 = 1296
And

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑒𝑎𝑟 𝑜𝑛 𝑓𝑜𝑢𝑟 𝑑𝑖𝑐𝑒 = 6 × 5 × 4 × 3
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑒𝑎𝑟 𝑜𝑛 𝑓𝑜𝑢𝑟 𝑑𝑖𝑐𝑒 = 360
So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑡𝑤𝑜 𝑑𝑖𝑐𝑒 𝑎𝑟𝑒 𝑠𝑎𝑚𝑒 = 1296 − 360
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑡𝑤𝑜 𝑑𝑖𝑐𝑒 𝑎𝑟𝑒 𝑠𝑎𝑚𝑒 = 936
Hence.

Sol: Here
𝑃(𝐴) =

2
3

𝑃(𝐵) = 𝑢
𝑢 =1−
𝑢=

3−2
3

𝑢=

1
3

𝑣 = 1−
𝑣=

2
3

1
5

5−1
5

𝑣=

4
5

𝑃(𝐷|𝐴) =

𝑃(𝐴 𝑎𝑛𝑑 𝐷)
𝑃(𝐴)

𝑃(𝐴 𝑎𝑛𝑑 𝐷) = 𝑃(𝐴)𝑃(𝐷|𝐴)
𝑤=

2
×𝑣
3

𝑤=

2 4
×
3 5

𝑤=

𝑃(𝐸|𝐵) =

8
15

𝑃(𝐵 𝑎𝑛𝑑 𝐸)
𝑃(𝐵)

1
𝑥= 6
𝑢
1
𝑥= 6
1
3
𝑥=

1
2

𝑦 = 1−𝑥
𝑦 =1−
𝑦=

2−1
2

𝑦=

𝑃(𝐹|𝐵) =

1
2

1
2

𝑃(𝐵 𝑎𝑛𝑑 𝐹)
𝑃(𝐵)

𝑃(𝐵 𝑎𝑛𝑑 𝐹) = 𝑃(𝐵)𝑃(𝐹|𝐵)
𝑧 =𝑢×𝑦

𝑧=

1 1
×
3 2

𝑧=

1
6

Hence.

Sol: Consider that leftmost digit is not zero, so the digit will be chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9.
Now the number will be odd if the rightmost digit of the number is from these odd
numbers 1, 3, 5, 7, 9.
The middle three digits will be chosen from these digits 0,1 ,2 3, 4, 5, 6, 7, 8, 9
So,
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑑𝑖𝑔𝑖𝑡 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 9 × 10 × 10 × 10 × 5
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5 − 𝑑𝑖𝑔𝑖𝑡 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 = 45000
Hence.

Sol: (a)
Total number of questions is 15 and the number of choices for each question is 3. So,
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡ℎ𝑒 𝑞𝑢𝑖𝑧 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑠𝑤𝑒𝑟𝑒𝑑 = 315
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡ℎ𝑒 𝑞𝑢𝑖𝑧 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑠𝑤𝑒𝑟𝑒𝑑 = 14348907
(b)
A grade can be obtained if the 14 questions are correct and one incorrect. The unanswered
question is also considered wrong.

Now 3 – 1 = 2 choices are wrong in each question. So, number of ways that the wrong
choice is selected from any of the 15 questions, for a question to be wrong is 15 × 2 = 30
Also, the unanswered question is also considered wrong.
So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 𝐴 𝑔𝑟𝑎𝑑𝑒 = 30 + 1
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 𝐴 𝑔𝑟𝑎𝑑𝑒 = 31
Hence.

Sol: (a)
There are two ways for the number plate. Letter first and then number or numbers first
and then letters.
Now, if repetition is allowed, all the four places for digits can be selected from the following
numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
And all the two places for letters can be selected from 26 alphabets.
So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑙𝑖𝑐𝑒𝑛𝑠𝑒 𝑝𝑙𝑎𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 2 × 10 × 10 × 10 × 10 × 26 × 26
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑙𝑖𝑐𝑒𝑛𝑠𝑒 𝑝𝑙𝑎𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 13520000
(b)
There are two ways for the number plate. Letter first and then number or numbers first
and then letters.
If repetition is not allowed, then the first digit is selected from the 10 digits which are 0,1 2,
3, 4, 5, 6, 7, 8, 9, second digit is selected from remaining 9 digits, third digit from remaining
8 digits, fourth digit from remaining 7 digits. First place for letter will be selected from 26
alphabets and the second place for letter will be selected from remaining 25 alphabets. So,
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑙𝑖𝑐𝑒𝑛𝑠𝑒 𝑝𝑙𝑎𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 2 × 10 × 9 × 8 × 7 × 26 × 25
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑙𝑖𝑐𝑒𝑛𝑠𝑒 𝑝𝑙𝑎𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 6552000

Hence.

Sol: (a)
𝑃(12,3) =

12!
(12 − 3)!

𝑃(12,3) =
𝑃(12,3) =

12!
9!

12 × 11 × 10 × 9!
9!

𝑃(12,3) = 12 × 11 × 10
𝑃(12,3) = 1320
(b)
𝐶(12,3) =

12!
3! (12 − 3)!

𝐶(12,3) =
𝐶(12,3) =

12!
3! 9!

12 × 11 × 10 × 9!
3! 9!

𝐶(12,3) =

12 × 11 × 10
3!

𝐶(12,3) =

12 × 11 × 10
3×2×1

𝐶(12,3) =

1320
6

𝐶(12,3) = 220
Hence.

Sol: (a)
When there is no restriction:
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 =

20𝐶10
2

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 =

184756
2

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 = 92378
(b)
When each team has one boy:
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 =

2𝐶1 × 18𝐶9
2

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 =

2 × 48620
2

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 = 48620
(c)
When all the boys are on same team:
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 = 2𝐶2 × 18𝐶8
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑡𝑒𝑛 − 𝑝𝑒𝑟𝑠𝑜𝑛 𝑡𝑒𝑎𝑚 = 43758
Hence.

Sol: (a)
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 8 𝑚𝑒𝑚𝑏𝑒𝑟𝑠 𝑡𝑜 𝑠𝑡𝑎𝑛𝑑 𝑖𝑛 𝑎 𝑙𝑖𝑛𝑒 = 8!
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 8 𝑚𝑒𝑚𝑏𝑒𝑟𝑠 𝑡𝑜 𝑠𝑡𝑎𝑛𝑑 𝑖𝑛 𝑎 𝑙𝑖𝑛𝑒 = 40320
(b)
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑐ℎ𝑜𝑜𝑠𝑒 𝑝𝑟𝑒𝑠𝑖𝑑𝑒𝑛𝑡 𝑎𝑛𝑑 𝑠𝑒𝑐𝑟𝑒𝑡...


Anonymous
Just what I needed…Fantastic!

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