In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
(3,1,1) toys=> 5C3 = 10 ways
(2,2,1) toys => (5C2*3C2)/2 = 15 ways
Total = 25?
@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
3^5 - 3(2^5 - 2) - 3
243 - 3 - 3(30)
150 ?
EDIT : 
sry identical he na boxes..this is for different boxes
sry identical he na boxes..this is for different boxes@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
Since the boxes are identical, only the groupings matter
Case 1: ( 3,1,1) => 5!/(3!1!1!2!) = 10
Case 2: (2,2,1) => 5!/(2!2!1!2!) = 15
Total = 25 ways ?
@sauravd2001 said:find the sum of squares and cubes of the root of eqaution x^3-2x^2+x-3?
a+b+c=2
ab+bc+ac=1
abc=3
( a+b+c)^2=a^2+b^2+c^2+2
=>a^2+b^2+c^2=4-2=2
( a+b+c)^3=a^3+b^3+c^3+3a(ab+bc+ac)+3b(ab+bc+ac)+3c(ab+bc+ac)-3abc
=>8=a^3+b^3+c^+6-9
=>a^3+b^3+c^=11
so 2+11=13
ab+bc+ac=1
abc=3
( a+b+c)^2=a^2+b^2+c^2+2
=>a^2+b^2+c^2=4-2=2
( a+b+c)^3=a^3+b^3+c^3+3a(ab+bc+ac)+3b(ab+bc+ac)+3c(ab+bc+ac)-3abc
=>8=a^3+b^3+c^+6-9
=>a^3+b^3+c^=11
so 2+11=13
@soham2208 said:Since the boxes are identical, only the groupings matterCase 1: ( 3,1,1) => 5!/(3!1!1!2!) = 10Case 2: (2,2,1) => 5!/(2!2!1!2!) = 15 Total = 25 ways ?
@sauravd2001 said:@pankaj1988 bhai 180??
25@rkshtsurana said:3^5 - 3(2^5 - 2) - 3243 - 3 - 3(30)150 ? EDIT : sry identical he na boxes..this is for different boxes
u r8...identical
@Messy_19 said:(3,1,1) toys=> 5C3 = 10 ways(2,2,1) toys => (5C2*3C2)/2 = 15 waysTotal = 25?
@rkshtsurana said:differentiate kro...u ll get pie/4 pe maximum...periodic function he ..so min at 5pie/4
periodic function kaise hua ye and pi/4 max. hai ye kaise pata chalega???
@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
5c3+5c2*3c2/2=10+15=25
Let a,b and c be the roots
Here a+b+c = 2
ab+bc+ca = 1
abc = 3
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
=> a²+b²+c² = (a+b+c)² - 2(ab+bc+ca) = 4 - 2 = 2
a³ + b³ + c³ = 3abc + (a+b+c)(a²+b²+c²-(ab+bc+ca))
= 3(3) + (2)(2-1)
= 11
ab+bc+ca = 1
abc = 3
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
=> a²+b²+c² = (a+b+c)² - 2(ab+bc+ca) = 4 - 2 = 2
a³ + b³ + c³ = 3abc + (a+b+c)(a²+b²+c²-(ab+bc+ca))
= 3(3) + (2)(2-1)
= 11
@sauravd2001 said:find the sum of squares and cubes of the root of eqaution x^3-2x^2+x-3?
@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
no arrangement only selection
3 1 1 = 5C3*2C1/2 = 10
2 2 1 = 5C2*3C2/2 = 15
total = 25
@sauravd2001 said:find the sum of squares and cubes of the root of eqaution x^3-2x^2+x-3?
a+b+c=2
ab+bc+ca=1
abc=3
(a+b+c)2=a2+b2+c2+2(ab+bc+ca)
=>a2+b2+c2=4-2*1=2
(a2+b2+c2)(a+b+c)=2*2=4
(a3+b3+c3+2(ab+bc+ca)-3abc=4
a3+b3+c3=4+3*3-2=11
@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
My take 25
3 1 1 5C3 = 10
2 2 1 5C2*3C2/2 = 15
25
@jain4444 said:no arrangement only selection 3 1 1 = 5C3*2C1/2 = 10 2 2 1 = 5C2*3C2/2 = 15 total = 25
Please explain your solution, I'm not getting. :/
There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____
(1) 5
(2) 21
(3) 33
(4) 60
(5) 27
(1) 5
(2) 21
(3) 33
(4) 60
(5) 27
@pankaj1988 said:In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
3 1 1 = 5C3 = 10
2 2 1 = 5C1 * 4C2/2! = 5 * 3 = 15
Total = 25
2 2 1 = 5C1 * 4C2/2! = 5 * 3 = 15
Total = 25
@pankaj1988 said:There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
6+5+4+3+2+1=21
@pankaj1988 said:There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
2)21
1G => 6C1
2G => 5C1
.
.
6G => 1C1
total = 21
1G => 6C1
2G => 5C1
.
.
6G => 1C1
total = 21
@pankaj1988 said:There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
Similarly, if 3 of the boxes have green balls, there will be 4 options.
If 4 boxes have green balls, there will be 3 options.
If 5 boxes have green balls, then there will be 2 options.
If all 6 boxes have green balls, then there will be just 1 options.
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
Similarly, if 3 of the boxes have green balls, there will be 4 options.
If 4 boxes have green balls, there will be 3 options.
If 5 boxes have green balls, then there will be 2 options.
If all 6 boxes have green balls, then there will be just 1 options.
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.