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In how many ways can Alok choose 15 colours from a box of 25 colours such that red, yellow, green, blue must always be included?

Option: 1

\frac{25!}{11!10!}


Option: 2

\frac{25!}{15!10!}


Option: 3

\frac{21!}{11!15!}


Option: 4

\frac{21!}{11!10!}


Answers (1)

Note the following:

  • The formula for the combination for the selection of the \mathrm{x} items from the \mathrm{y}different items is \mathrm{=^{y}C_{x}=\frac{y!}{x!\left ( y-x \right )!}}

  • The restricted combination for the selection of the \mathrm{r} items from the \mathrm{n}different items with \mathrm{k} particular things always included is \mathrm{=^{n-k}C_{r-k}}

Since 4 colours must always be included, the following is evident.

  • The number from which the restricted combination is to be made is

          =\mathrm{n-k=25-4=21}.

  • The number with which the restricted combination is to be made is

          \mathrm{=r-k=15-4=11}

Therefore, the required restricted combination is

=\mathrm{^{n-k}C_{r-k}}

\mathrm{=^{21}C_{11}}

=\frac{21!}{11!10!}

Posted by

Sumit Saini

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